Sir, beautifully & patiently explained. You have an ocean of patience which is the prime requirement of any teacher Even a student weak in Maths will easily understand if u take a class Thanks
Nice geometric solution! I got there with tangents function. If you complete the right-angled triangle in the bottom left, with height h and base p+2d (were d is the length AD) then you see there are three right-angled triangles and by considering each one in turn that: tan 60 = (p+2d)/h = sqrt(3) tan 45 = (p+d)/h = 1 tan (45-x) = p/h Multiply both sides of the second equation by two and subtract the first equation from it and you have 2-sqrt(3) = p/h Substitute this into the third equation and you get: x = 45 - tan-1 (2-sqrt(3)) which is 30 degrees.
Let's say that you are required to solve by construction only, no trigonometry. Eastern Brown's solution adapts as follows. Complete the right triangle and make use of the known ratios of sides for the 60°-30°-90°, 45°-45°-90°, and 75°-15°-90° right triangles. The smallest right triangle has p/h = (2-√3) = (√3-1)/(√3+1) which matches the 75°-15°-90° right triangle. The smallest angle, 15°, needs to be subtracted from 45° to get x = 30°.
Sir, you are truly the Sherlock Holmes of trigonometry. You assemble the evidence point by point, and arrive at the solution step by step. It is making mathematics good fun! Thank you.
I solved by using trignometry and the law of sines for triangles. With reference to your diagram, let AD=BD= a, and let us designate the length AC=b. Then, for the triangle ACD, a/sinx = b/sin45 ==> sinx = a/(b*sqrt 2). Also for the triangle ACB, 2a/sin(x+15) = b/sin 30 ==> sin(x+15) = a/b = sqrt 2*sinx. Therefore, sinx*cos15+cosx*sin15 = sqrt 2*sinx. If we divide this equation by sinx, it can be reduced to cos15 + cotx*sin15 = sqrt 2 ==> cotx = (sqrt 2 - cos15)/sin15, or tan x = sin15/(sqrt 2 - cos15). The right hand expression can be calculated as 0.57735026919, which happens to be the tan 30 degrees.
True, but the video shows the fundamental or the manual way ehe, even for those who haven't learned trigonometry still can solve the problem... Nice video ☺️
In ∆CDB, angle B + 15° = 45° [exterior angle is the sum of interior opposite angles] Therefore angle B = 30° Let AD = DB = a and AC = b In ∆ABC by law of sine 2a/sin(x+15°) = b/sin30° = b/(1/2) = 2b => sin (x+15°) = a/b --------(1) In ∆ ADC a/sin x = b/sin45° = b/(1/√2) = b√2 => √2 sin x = a/b ---------(2) From (1) and (2) √2 sin x = sin (x+15°) √2 sin x = sin x cos 15° + cos x sin 15° (√2-cos 15°) sin x = cos x sin15° sin x / cos x = sin 15° /(√2-cos 15°) tan x = sin 15° / (√2-cos 15°) tan x = 1/√3 ( pl. refer Note below) = tan 30° Thus the unknown angle x = 30° Note: sin 15° = sin (45°-30°) = sin 45° cos 30° - cos 45° sin30° = [1/√2] [ √3/2 - 1/2] = (√3-1)/2√2 Similarly cos 15 = (√3+1)/2√2 √2-cos 15 = √2 - (√3+1)/2√2 = (3-√3)/2√2 = √3(√3-1)/2√2 = √3 sin 15 Hence sin 15° / (√2-cos 15°) = 1/√3
3:13 to get angle PDC you don't need the 120 degree. Just use the exterior angle theorem again in the trangle PDC with angle BPD=30degree as the exterior angle, and angle PCD=15degree so angle PDC=30-15=15degree. This is a good question that looks difficult but once the additional lines are drawn it becomes straightforward.
Don't know what your age or former occupational title has to do with it. And you "guessed" it correctly? There's no guessing as you couldn't possibly know if you "guess" was correct until someone else DEMOSTRATED it to be correct. Ugh You must have been an excellent engineer. SMH
*Trigonometric solution:* Drop a perpendicular from C to meet the base line at E. Let CE= a and ∠ECA= θ. Since ∠DCE=45°, we have ED=a, and since ∠EBC= 30°, we have EB=a✓3. This gives AD= (✓3-1)a and EA=ED-AD=(2-√3)a. Now Tanθ=EA/CE = 2-√3, which gives us θ=15°. Hence X=30°.
I used the law of sines and cosines. Since line AD = DB it's easier to make an assumption about the length and then we can workout CD using the sine rule and CA using the cosine rule then we can use the sine rule again to find the value of X. But thank you, this was a good exercise.
Exactly.. and using this method, you can find that 2x = 60 in less than 1.5 minutes. That's way more than the allotted time for solving a geometry problem like this in CAT exam. We usually have 25-30 seconds for solving such problems in CAT exam.
Well, there are many ways of solving this beautiful problem!! 1) using elementary geometry. 2) using sine rule 3) using m-n theorem Using the third method gives u the answer in just 1step...😎😎😉😉 Anyways, thanks for sharing! Love from India!!
M-N theorem is very handy and interesting. Went through its proof. (m+n)cot(theta)=m.cot(alpha)-n.cot(beta). In this problem AD=m DB=n; m=n Theta=135 Alpha = X Beta = 15 Sin(15) = (√3-1)/(2.√2) Cos(15) = (√3+1)/(2.√2) CotX = √3 X = 30. Very nice approach. Today i learnt m-n theorem. Thank you Kusuma I used the second approach
Interesting solution. Another way is drawing CH and proving that A and H are one point, because the other two possible positions of H lead to contradictions with the triangle CHB.
This seems to be one of a number of similar problems that are special situations. We are given a 45 degree ray from the midpoint D of AB and an intersecting ray from A to C. The problem is to find angle ACD. The method of solution only works if the point C is chosen so that angle DCB is 15 degrees or alternatively angle DBC is 30 degrees..
Intuitively, I figured that the bifurcation of the base created a triangle ADC which you could map to ABC by a reflection across AC, and a rotation around A, followed by a dilation. This would show that angle BAC is the same angle as CAD. The measure of angle BAC I got deriving angle DBC from 180 - 45 = 135, and then 180 - 15 - 135 = 30. As it turns out, this leads to the correct solution, namely x = 30 degrees. But I don't know if this is a fluke :)
It's over 30 years since I took geometry in university (a Euclidian and non-Euclidian course). I thought it was not possible to find X. I also misinterpreted the double hashes as meaning AD and DB were parallel :D So I didn't get far. I was amazing by the explanation. Such a smart tactic. Thanks for taking the time.
an easier solution i think would be to just draw a line say CE parallel to AD now using alternate interior angle property we can conclude that angle ADC = angle DCE. Now since the line CE is also parallel to DB therefore we can conclude using alternate angle property that angle DBC = angle BCE. Now angle DCE= angle DCB + angle BCE = 45. Therefore, 15 + BCE = 45, therefore BCE = DBC = 30
All angles except ones on left side are known. The ones on left side are x and 135-x. Use law of sines to relate ratio of shared side length to length of congruent side. Solve for x. x = invtan(sin135(sin30/sin15+cos135)^-1) = 30.
Great puzzle- I needed the DP portion to start before I could solve it. Thanks for providing such good content and a mix of easier and more difficult problems.
Very nice approach. I solved it using sine rule. sin(15+x)/2sinx=sin30/sin45=>sin(15+x)/sinx=sin45/sin30. Putting x=30 or 15+x=45 gives the same result x=30. Hence x=30 deg.
In 1970, my Maths teacher at Whitchurch high school in Cardiff, Mr Smth, was so good he helped me to learn how to do this stuff and I am forever grateful. I still haven't found a use for the integration of 1 + Tan squared (x) though. :)
@@Z7youtube It's in my comment. Using formula for sine of a difference. sin(135°-x) = sin(135°)cos(x)-cos(135°)sin(x) = (√2/2) cos(x) + (√2/2)sin(x) We used the fact sin(135°)=sin(45°)=√2/2 and cos(135°)=-cos(45°)=-√2/2
Yes, I found “X”. It was hiding just to the lower right of “C” at the top. It was pretty easy to find. Especially since it was written in red! 😊 Just kidding. I never thought I would use calculus or trigonometry much when I was in high school. Then I got into machining and use it almost daily! Pay attention kids this IS important!
Notice that sin(x)/AD = sin(45)/AC and sin(x+15)/(2*AD) = sin(30)/AC. Combine to get 2*sin(x)/sin(x+15) = sin(45)/sin(30). Upon expansion of sin(x+15) and some algebra, we get x = 30 degrees.
Great puzzle! I didn't find this excellent geometry solution. Instead I dropped a perpendicular from C to point E, then used trig to get CE, EA and EB (after setting AD and BD=1). Then I used Pythagoras to get CB=sqrt 2. Then the Law of Sines to get sin x=1/2. Thanks PreMath!
@@hussainfawzer Hello Thanks for the question, I left out a lot of steps and also erred in saying CB=sqrt2. Sorry for that (CA=sqrt2 is correct). To get EA then CE let EA=y and AB=1 (or any number but 1 makes the calculating easier) Since triangle CED is a right triangle with a 45 degree angle, CE=ED=1+y. EB also =2+y Since angle CEB is defined as a right angle and CBE (as shown by PreMath) is 30 degrees, CEB is a 30-60-90 triangle and CB will be twice CE or 2+2y So by Pythagoras CE^2+EB^2=CB^2 or (1+y)^2+(2+y)^2=(2+2y)^2 Simplify and get 2y^2+2y-1=0 and so y=(sqrt3-1)/2 and EC=(sqrt3+1)/2 I hope that answers your question, Hussain. Now one can get CA=sqrt2 from Pythagoras. And looking at triangle CAD, the law of sines gives sinx/1=sin45/sqrt2 and sinx=1/2.
@@waheisel That was beautiful math. It reminded me of a lot of stuff. Ty I used to love math and physics with a dream to be get into cosmology. Due to personal reasons I'm now soon becoming a doctor. It's good but I miss math and physics so badly 😅
@@yuda4626 Thanks for the kind comment Ty. When I was in school I also liked math, physics, and astronomy. I wasn't nearly good enough to make any of those a career. I also became a doctor. Even though you won't use so much math and physics in your career you can still enjoy your daily PreMath puzzle! Best Wishes
Question? I am very rusty in terms of geometry. The last course taken was during the 1984-1985 academic school year. From the diagram above might we say that 180 = 30 + ( x + 15 ) + Angle CAD and 180 = 45 + x + Angle CAD and we know by the rule of supplementary angles 180 = 45 + Angle CDB. So, by this rule, we know that Angle CDB has a value of 135. And, by the exterior angle rule given, 135 = x + Angle CAD. Based upon this information, drawn from the given, is it possible that angle x equals 25 and Angle CAD equals 110. Those numbers might be arising in my vision from the CBD smoke seeping in through the walls in this apartment unit; because, I do not smoke and do not physically tolerate it well. A year or so after taking geometry, linear algebra taught that based upon the information given within a problems statement more than one "answer" might exist in the solution-space. And, that might be what is occurring here. If, I am not mistaken. So, we both might have settled on accurate solutions. I still a little puzzled by that external angle rule. Unless, Angle CAD can also have a possible measure of 105. Maybe, that is why it is labelled CAD?😄
Before these things used to be a headache for me in class. I did it cause I was forced to and did the minimum. Now it’s my entertainment and I can’t get enough of it. I rather watch this than a series.
After watching few similar videos able to solve this one in mind. It's mostly in finding/drawing the correct isosceles, equilateral or congruent triangles with the equal length sides.
I admit I got stuck after step one. designating point P was a clever trick, but only worked out by chance, due to the particular properties of the two triangles. After it turned out that the vertex in D of APD is 60 degrees, the rest was easy!
Fantastic and clever solution! I couldn't made it that way! I have another aprroach: 1. Draw the circle that contains point A,B,C. To do so, draw 2 perpendicular lines at the midpoind of AB and BC, these two lines meet at the center O. WE notice that AOB is the angle at the center and the angle ACB (= angle x +15 degrees) is the angle at the circumference. 2. Demonstrate that the angle at the center AOB is equal 90 degrees. Then, we can find the value of the angle ACB = 1/2 AOB= 45 degrees. 3. The angle x= angle ACB - 15 = 45-15= 30 degrees.
@@emaceferli726 Yes, I can. Just label M and N the midpoint of AB and then AC. Draw 2 perpendicular lines to AB, and AC at these points which meet at the center O. Then we have OC=OA=OB, therefore we can draw a circle from the center O with the radius R=OC=OA=OB. It is easy to see that the angle CBA=30 degrees (CMA is the exterior angle=45degrees). The angle CBA is the angle at circumference of which COA is the angle at the center, so COA= 2 CBA=60 degrees----> the triangle COA is an equilateral one. Let's say I is the midpoint of OA. The height from the vertex C to I is also the bisector of the angle ACO (=60 degrees). Because the angle AMC=OMC=45 degrees so CM is the bisector of the angle OMA, so C,I,M are colleniar. Thus the angle x= ACM= 30 degrees
I got it. My sister told me the triangle CDP is an isosceles because angle PCD and PDC have the same 15 degree. Therefore, PC = PD. The triangle APD is equilateral, hence PC = PA. So angle PCA and PAC are 45 degree each. Consequently, angle ACD is 30 degree. That’s awesome!!! Thank you very much. I had learned geometry 50 years ago. I have to confess this brings back a lot of memories.
Beautiful solution to the problem. Very much in the domain where math is art itself. I have opted for a cruder way (PS: I am an engineer). I dropped a perpendicular instead and used expressions for tan. Got two equations and soleved simultaneously to get the answer.
Another flash of brilliance from the maestro, that was a superb demonstration, I'd never have got that, I like the way in many of your marvels that you add cunning lines that reveal the pathway, it's a joy to watch. Thanks again 👍🏻
You could've also just found angle BDC as it's a linear pair=135 and then as cdb is a triangle with angle sun property you can find angle b as 30 and then angle acd equals dbc so angle acd is 30
THE SHORT: If you could draw DP so that DP=DB, on paper, you could also just as easily draw AP so that AP=AD=CP=PD, on paper... all anchored by the given angles of ADC & DCB. THE LONG: Imagine vertices ABC are 3 stars in a system, with D being a star exactly between star A and star B (let's say with Hubble, we're easily able to measure that the star D is exactly half way, in light years, between stars A & B, all three nicely aligned). From ANY of these given stars, to any other star (in the ABCD set of star system), as well as, angles 15 & 45....nothing was given, could have been given, via assumption or "eye balling".... Everything came about through precise measurements via parallax + trigonometric, standard Candle method, etc. Take one step back and ask yourself: --- what is the astronomical (if we took ABCD to be a set of 4 stars in the Milky Way) or geometric logic that said you could, for example, draw a precisely KNOWN line segment, like DP, from one vertex of BDC, to the opposite side, BC, in such a way that DP is exactly equal to one side, DB.... but NOT be able to do a similar thing, say, AP? After all, AP=AD=DB, DB=DP, and AD=DB.... all anchored by the given angles of 15 (DCB) & 45 (ADC). For example, "P" can NOT be in any other place, along BC, such that AD is not equal to AP (due to the dictates of the angles 15 and 45, as given). In other words, there's neither "wiggle room" for angles ADC & DCB, on the one hand, nor for point "P" along BC such that AP=AD=DB=DP is NOT true, on the other.
I took Geo Trig last school year and felt very smart knowing how to solve everything. However when I saw this, it was not the case. Thanks for explaining this very thoroughly. It was very fun and interesting to learn how to solve this Geo Challenge. Please make more!
Thanks for the excellent explanation. I stared at this problem for several minutes before coming up with the answer. It seemed obvious to me that angle X should be equal to angle ABC and ABC was obviously 30. based on the given angles, but while I could prove angle ABC couldn't figure out how to prove angle X.
Another method is to use sine theorem and a trig identity to prove sin x /cos x = (sqrt 3)/3 = tan x. So x =30. Less head scratching to figure out how to draw auxillary lines.
Hi, you could have separated triangle cdb and joined AD and DB by rotating clock wise triangle CDB. Making points A and B to be the same point. And you can get the answer in 2 steps. As rotating would create an isosceles triangle. Try it out
Rotating does not create an isosceles triangle. In fact, in the general case, the rotation is more likely to create a 4 sided polygon. In this particular case it creates a triangle, but not an isosceles one. To think this through, think about point C being split by the rotation, where C1 is the original point C, and C2 is created by the rotation. In this case, because angle ADC is 45 degrees and angle CDB is 135 degrees, after the rotation the line segments C1D and DC2 are collinear forming a straight line, with D in the center, since C1D and DC2 are the same length (but since they form a straight line they are not two sides of an isosceles triangle). So one side of the resulting triangle is C1C2, with the other two sides being AC1 and BC2 (A and B are the same point after the rotation). One of the angles in this new triangle is the 15 degrees from angle DCB prior to rotation, another angle is X degrees, formed from angle ACD prior to rotation (now AC1C2 after rotation). We know from the solution presented that X is not 15 degrees but is a value (trying not to spoil) such that none of the angles in the triangle formed by the rotation are the same.
i guess in these kind of problems only the one who created them can solve them, or you can go through a long journey of trying all possible methods and theoremes to finally conclude it
Clearing away the cobwebs. 55+ years I was solving problems like this as part of a submarine fire-control problem (I.E. what angle to set the torpedo to intersect with the surface target) We had a mechanical analog computer, however, if it failed, you had to know how to do it by hand. Thank you. Narragansett Bay
I was just wondering how will you use that information, that those two parts of bottom line is equal, and I'm speechless how beautifully you created a triangle out of it and used that fact... I knew that equal sides of triangle subtend equal opposite angles in a triangle, but i wasn't smart enough to use that fact... Thank you sir, love from India..!
During my jee preparation (engineering entrance exam of India) i studied a theorem called m-n cot theorem, and really it gave the answer in just 5-10 seconds :) 😁 BTW great explanation sir😊
will it work out if you simply express all sides with sine rule in a way that the sides cancel out and only an equation for sin(x) remains, leaving an answer of sin(x)=1/2, which due to that 45 degree angle makes 30 deg the only solution?
I think the question may be solved by applying basic triangle rules. I tried by following method- 1. By external angle theorem: angle DBC = 30° 2. In triangle DBC, if length of DB is 'a' (against 15°), then CD is 2a (against 30°) 3. Now, in triangle CAD, AD = a, CD = 2a 4. Angle ACD = x, Angle ADC = 45° = y, Angle CAD = z 5. Angle z = 2 (Angle x) [because CD = 2 (AD)]. 6. As, Angle x + y + z = 180° x + 45 + 2(x) = 180° 3x = 180° - 45° = 135° x = 135°/3 x = 45°
There is no rule in a triangle as you wrote "if length of DB is 'a' (against 15°), then CD is 2a (against 30°)". By this info you can just write: CD>AB.
Is it only me who thought that if AD=DB, CD is the median line of the triangle in C, thus meaning in cuts the angle in half? And if DCB is 15, ACB would be 30?🤔🤔
Actually, that is not always the case, as shown in this problem here. The median of a triangle in a given vertex is not always equal to the bisector of the relative angle. They're the same in some cases, for example if the triangle is equilateral, it is true for alla the vertices.
You're mistaken between median line and angle bisector. This is proven false in this example so if it cit ACB in middle = ACD = DCB =15° which is not the case
Solved using a very long method. A shorter way to do it is to extend CD to P so that CD=DP. Triangle DBP can be proved congruent to triangle CAD and angle X will be equal to angle DPB. DPB will be first calculated to be 30 degrees.
Me too. Then, as tan 15°=2 - √3 (use the compound angle formula to evaluate tan 15°=tan(45°-30°) to get this), we get x=45°-15°=30°. I didn't spot the geometric solution until I saw the construction of AP in the video.
@@PreMath And here the details for all others: Draw the height from point C to the A-B line. Call the foot point as E. In my way, I also calculated the angle at point B first, i.e. 30°. The triangle BCE shows a 30°/60°/90° triangle. So when BC has a length of 2 units, then CE has lenght 1 unit and BE is sqrt(3) units long. CD then is sqrt(2) units long because of the 45° angle and DE is 1 unit long (40°/45°/90° triangle CDE). Lengt BD = BE - DE = sqrt(3) - 1. Because AD = BD =sqrt(3)-1 units then AE = DE - AD = 1 - (sqrt(3) - 1) = 1 - sqrt(3) + 1 = 2 - sqrt(3). Focus on the right triangle ACE where the angle ACE can be calculated as tan(ACE) = AE/CD = (2 - sqrt(3))/1 = 2 - sqrt(3). And important to know: tan(15°)=2-sqrt(3) is also a special value (can be easily proved on a 75°/75°/30° isosceles triangle). The actual result x is just the difference of the angle BCE (45°) and ACE (15°) in this case.
could use trig by drawing a perpendicular line from C to P to form right triangle CPA. Triangle CPB is a 30-60-90 right triangle. label the sides as 1, sqrt 3 (or 1.732) and 2 let's label angle CPA as 'y' . y + x =45 degrees and you now have and 45-90-45 with sides 1, 1, and sqrt 2 degree right-triangle line PD =1 and CD= sqrt 2 . Using ASA (15 degree, sqrt 2 and 135 degree) for triangle CDB gives 0.732 for DB , but DB = AD thus PA = 1-0.732 = 0.268. Since the right triangle CPA has the sides 1, 0.268 then using SAS (1, 0.268, and 90 degree) gives 15 degrees for angle y, but y+x=45 degree hence x=30 degree
30 degrees ( since ACB angle is 45 degrees). Angle ABC is 30 degrees, draw the line DE parallel to CB ( which crosses AC in E). Length of DE is 1/2 of CB, since AD = DB. Then draw two lines from D and E, each perpendicular to CB. Denote that the only thing which needed is to know that in the right triangle with angles 30/60, the side opposite to 30 degree angle is 1/2 of side opposite to 90 degree angle, and side opposite to 60 degree angle is √3/2 of it. For that no nowledge of trigonometry is required, only Pythagorean theorem.
It was an easy puzzle! You could've solved it with a much easy technique. ∠CDA + ∠CDB=180°[Linear Pair] . So, ∠CDA is given as 45°. Then, 45°+ ∠CDB=180°. ∠CDB=135°. ∠B+∠CDA+∠DCB=180°[Angle sum property of a triangle]. Then if we construct a similar congruent triangle to CAD say CAE. we get x+x+15°+45°+45°=180°[angle sum property of a triangle]. we get the value of x as 45°. So, this was a more easy technique😀😀
I remember a short cut, but I'm old and don't know where I learned it. Basically if the line segments are the same then the angle x = the angle CBD. so all you have to do is solver for CBD. Angle ADC may have to be 45 though(it might have to do with trig). I don't remember.
I used sin rule. Sin15/DB = sin30/CD. I get DB = 0.518CD. and then sinx/DB = sin(135-x)/CD. I substituted DB = 0.518CD into the second sin rule equation. CD got cancelled out and after using some trigo identities, I managed to solve for x as the answer given. Good qns!
@@starpupil1843 u r welcome! My weakness involves me not knowing what line to draw or what to substitute (in calculus) to get to my solution faster. I deal with what I see and what I can find base on what I see. Hahs
Extend line CA and obtain exterior angle (180-x-15) =x+ ang CAD thats one equation. Another eqn is using triang.CAD : x + Ang CAD =135. We have two eqns and two unknowns which we can solve to get x=30
This question is actually very easy Draw a perpendicular to line AB, let the point be H Assume length AD as l Write Tan45 and tan30. Solve for AH and CH in terms of l Find CH/AH, this is tan(45-x) We get tan(2-√3) = 45-x
Sir, beautifully & patiently explained. You have an ocean of patience which is the prime requirement of any teacher
Even a student weak in Maths will easily understand if u take a class
Thanks
Wow!
Thank you for your nice feedback! Cheers!
You are awesome Pracash😀
@@PreMath I like the way you quote a theorem for every step.
You can use tringonometry too
Then ,It will be so short
It is very long method
I think you should work on more specified way to solve problem
Nice geometric solution!
I got there with tangents function. If you complete the right-angled triangle in the bottom left, with height h and base p+2d (were d is the length AD) then you see there are three right-angled triangles and by considering each one in turn that:
tan 60 = (p+2d)/h = sqrt(3)
tan 45 = (p+d)/h = 1
tan (45-x) = p/h
Multiply both sides of the second equation by two and subtract the first equation from it and you have 2-sqrt(3) = p/h
Substitute this into the third equation and you get:
x = 45 - tan-1 (2-sqrt(3))
which is 30 degrees.
Correct
Let's say that you are required to solve by construction only, no trigonometry. Eastern Brown's solution adapts as follows. Complete the right triangle and make use of the known ratios of sides for the 60°-30°-90°, 45°-45°-90°, and 75°-15°-90° right triangles. The smallest right triangle has p/h = (2-√3) = (√3-1)/(√3+1) which matches the 75°-15°-90° right triangle. The smallest angle, 15°, needs to be subtracted from 45° to get x = 30°.
superb
👍👍👍
Sir, you are truly the Sherlock Holmes of trigonometry. You assemble the evidence point by point, and arrive at the solution step by step. It is making mathematics good fun! Thank you.
This is geometry. Simple euclidean theorems taught in 6th and 7th. Not trigonometry.
I was drawing angles after angles, but no joy. Your explanation is great! Thanks!
I solved by using trignometry and the law of sines for triangles. With reference to your diagram, let AD=BD= a, and let us designate the length AC=b. Then, for the triangle ACD,
a/sinx = b/sin45 ==> sinx = a/(b*sqrt 2). Also for the triangle ACB, 2a/sin(x+15) = b/sin 30
==> sin(x+15) = a/b = sqrt 2*sinx. Therefore, sinx*cos15+cosx*sin15 = sqrt 2*sinx. If we divide this equation by sinx, it can be reduced to cos15 + cotx*sin15 = sqrt 2 ==> cotx =
(sqrt 2 - cos15)/sin15, or tan x = sin15/(sqrt 2 - cos15). The right hand expression can be calculated as 0.57735026919, which happens to be the tan 30 degrees.
Very good
I solve it same way ❤️❤️
True, but the video shows the fundamental or the manual way ehe, even for those who haven't learned trigonometry still can solve the problem... Nice video ☺️
Except that trigonometry applies only to right triangle.
@@dilasgrau6433 Nope. Law of sines and law of cosines is also considered trigonometry. Anything that uses trig ratios is considered trigonometry.
In ∆CDB, angle B + 15° = 45° [exterior angle is the sum of interior opposite angles]
Therefore angle B = 30°
Let AD = DB = a
and AC = b
In ∆ABC by law of sine
2a/sin(x+15°) = b/sin30°
= b/(1/2)
= 2b
=> sin (x+15°) = a/b --------(1)
In ∆ ADC
a/sin x = b/sin45°
= b/(1/√2)
= b√2
=> √2 sin x = a/b ---------(2)
From (1) and (2)
√2 sin x = sin (x+15°)
√2 sin x = sin x cos 15° + cos x sin 15°
(√2-cos 15°) sin x = cos x sin15°
sin x / cos x = sin 15° /(√2-cos 15°)
tan x = sin 15° / (√2-cos 15°)
tan x = 1/√3 ( pl. refer Note below)
= tan 30°
Thus the unknown angle x = 30°
Note:
sin 15° = sin (45°-30°)
= sin 45° cos 30° - cos 45° sin30°
= [1/√2] [ √3/2 - 1/2]
= (√3-1)/2√2
Similarly
cos 15 = (√3+1)/2√2
√2-cos 15 = √2 - (√3+1)/2√2
= (3-√3)/2√2
= √3(√3-1)/2√2
= √3 sin 15
Hence
sin 15° / (√2-cos 15°) = 1/√3
The hardest part of solving geometry problem is to find a point like P.😂😂😂
I agree.
Yes bro 😂😂😂
3:13 to get angle PDC you don't need the 120 degree. Just use the exterior angle theorem again in the trangle PDC with angle BPD=30degree as the exterior angle, and angle PCD=15degree so angle PDC=30-15=15degree.
This is a good question that looks difficult but once the additional lines are drawn it becomes straightforward.
I am 72 and a retired electrical engineer. I guessed it correctly however, your step by step explanation was excellent. Thank you.
Don't know what your age or former occupational title has to do with it. And you "guessed" it correctly? There's no guessing as you couldn't possibly know if you "guess" was correct until someone else DEMOSTRATED it to be correct. Ugh You must have been an excellent engineer. SMH
Same. It was fu. Reliving the old steps
*Trigonometric solution:*
Drop a perpendicular from C to meet the base line at E. Let CE= a and ∠ECA= θ. Since ∠DCE=45°, we have ED=a, and since ∠EBC= 30°, we have EB=a✓3. This gives AD= (✓3-1)a and EA=ED-AD=(2-√3)a.
Now Tanθ=EA/CE = 2-√3, which gives us θ=15°. Hence X=30°.
Great!
Thank you for your nice feedback! Cheers!
You are awesome Hari.😀
Very well thought out. Really nice alternative solution
very nice - designating point "P" was critical, and unfortunately I didn´t see that on my own - jajaja
So nice of you Charles.
Thank you for your feedback! Cheers!
You are awesome.
Keep rocking😀
@KWC Coin Try backing up and muscling it through with law of sines and law of cosines.
Yeah
Finding out what construction to make is always the hardest part ... I can never figure it out.
ua-cam.com/video/Z67zK4B6cfU/v-deo.html😊😊
I used the law of sines and cosines. Since line AD = DB it's easier to make an assumption about the length and then we can workout CD using the sine rule and CA using the cosine rule then we can use the sine rule again to find the value of X. But thank you, this was a good exercise.
Yes - sine rule - that is how I would have done it.
Exactly.. and using this method, you can find that 2x = 60 in less than 1.5 minutes. That's way more than the allotted time for solving a geometry problem like this in CAT exam. We usually have 25-30 seconds for solving such problems in CAT exam.
How you did it? I used cosine and sine law but took like way more time than 1.5min
Well, there are many ways of solving this beautiful problem!!
1) using elementary geometry.
2) using sine rule
3) using m-n theorem
Using the third method gives u the answer in just 1step...😎😎😉😉
Anyways, thanks for sharing!
Love from India!!
M-N theorem is very handy and interesting. Went through its proof.
(m+n)cot(theta)=m.cot(alpha)-n.cot(beta).
In this problem
AD=m DB=n; m=n
Theta=135
Alpha = X
Beta = 15
Sin(15) = (√3-1)/(2.√2)
Cos(15) = (√3+1)/(2.√2)
CotX = √3
X = 30.
Very nice approach.
Today i learnt m-n theorem.
Thank you Kusuma
I used the second approach
@@sandanadurair5862
Nice 👍!!
Thanks for sharing! Cheers!
You are awesome Kusuma😀
Love and prayers from the USA!
When do you use the m-n theorem?
How to solve using m-n theorem ?
Interesting solution. Another way is drawing CH and proving that A and H are one point, because the other two possible positions of H lead to contradictions with the triangle CHB.
You explained way better than my last years Math teacher did! Thanks for helping me clearly understand this process!
You're very welcome Aakash
Thank you for your nice feedback! Cheers!
You are awesome.😀
@@PreMath Thanks.
This seems to be one of a number of similar problems that are special situations. We are given a 45 degree ray from the midpoint D of AB and an intersecting ray from A to C. The problem is to find angle ACD. The method of solution only works if the point C is chosen so that angle DCB is 15 degrees or alternatively angle DBC is 30 degrees..
Intuitively, I figured that the bifurcation of the base created a triangle ADC which you could map to ABC by a reflection across AC, and a rotation around A, followed by a dilation. This would show that angle BAC is the same angle as CAD. The measure of angle BAC I got deriving angle DBC from 180 - 45 = 135, and then 180 - 15 - 135 = 30. As it turns out, this leads to the correct solution, namely x = 30 degrees. But I don't know if this is a fluke :)
Not a fluke. The method presented os a very long winded approach to achieve the same result.
It's over 30 years since I took geometry in university (a Euclidian and non-Euclidian course). I thought it was not possible to find X. I also misinterpreted the double hashes as meaning AD and DB were parallel :D So I didn't get far. I was amazing by the explanation. Such a smart tactic. Thanks for taking the time.
Great job demonstrating how one can use simple facts of geometry and a little deduction to reduce the problem to a simple math problem.
Well done, I didn't think it was possible to solve this problem without calculating a single length!
You can absolutely solve just based upon sum of enclosed angles without the unnecessary analytic geometry
You cant calculate a single length. There isn't enough info.
@@anonymousman1282 true, but ratio of length is all you would need, if you would want to do it another way.
an easier solution i think would be to just draw a line say CE parallel to AD now using alternate interior angle property we can conclude that angle ADC = angle DCE. Now since the line CE is also parallel to DB therefore we can conclude using alternate angle property that angle DBC = angle BCE. Now angle DCE= angle DCB + angle BCE = 45. Therefore, 15 + BCE = 45, therefore BCE = DBC = 30
thats not x
All angles except ones on left side are known. The ones on left side are x and 135-x. Use law of sines to relate ratio of shared side length to length of congruent side. Solve for x. x = invtan(sin135(sin30/sin15+cos135)^-1) = 30.
Great puzzle- I needed the DP portion to start before I could solve it. Thanks for providing such good content and a mix of easier and more difficult problems.
Glad to hear that!
Thank you for your nice feedback! Cheers!
You are awesome Jay.😀
Love and prayers from the USA!
None of that was DP math...
Please make this kind of geometry video more...and make pdf including all their rules and strategy
Very nice approach. I solved it using sine rule. sin(15+x)/2sinx=sin30/sin45=>sin(15+x)/sinx=sin45/sin30. Putting x=30 or 15+x=45 gives the same result x=30. Hence x=30 deg.
Great Satyanarayan
Thank you for your feedback! Cheers!
You are awesome.😀
Trigonometry works only on right angled triangle, right?
In 1970, my Maths teacher at Whitchurch high school in Cardiff, Mr Smth, was so good he helped me to learn how to do this stuff and I am forever grateful. I still haven't found a use for the integration of 1 + Tan squared (x) though. :)
Excellent little problem, thank you!
Can we solved simpler, I think, using law of sines.
In triangle BCD, sin
Cool! Many ways to solve this problem.
Thank you for your nice feedback! Cheers!
You are awesome Vsevolod.😀
I use the same way to slove this question.
can u please plz tell me only what does sin(135-x) equal? i didn't get it from ur comment
@@Z7youtube It's in my comment. Using formula for sine of a difference.
sin(135°-x) = sin(135°)cos(x)-cos(135°)sin(x) = (√2/2) cos(x) + (√2/2)sin(x)
We used the fact sin(135°)=sin(45°)=√2/2 and cos(135°)=-cos(45°)=-√2/2
@@vsevolodtokarev oh ok i didn't know about the formula for sine of a difference , tysm!
Yes, I found “X”. It was hiding just to the lower right of “C” at the top. It was pretty easy to find. Especially since it was written in red! 😊 Just kidding. I never thought I would use calculus or trigonometry much when I was in high school. Then I got into machining and use it almost daily! Pay attention kids this IS important!
Hey, can I ask more about the field in which you work?
@@Manifestingqueen1111 sure
Kudos to anyone who can solve problems like this, but my hat is really off to those who can CREATE problems like this.
Notice that sin(x)/AD = sin(45)/AC and sin(x+15)/(2*AD) = sin(30)/AC. Combine to get 2*sin(x)/sin(x+15) = sin(45)/sin(30). Upon expansion of sin(x+15) and some algebra, we get x = 30 degrees.
Great puzzle! I didn't find this excellent geometry solution. Instead I dropped a perpendicular from C to point E, then used trig to get CE, EA and EB (after setting AD and BD=1). Then I used Pythagoras to get CB=sqrt 2. Then the Law of Sines to get sin x=1/2. Thanks PreMath!
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome William😀
How did you fin CE ?
Could you please explain
@@hussainfawzer Hello Thanks for the question, I left out a lot of steps and also erred in saying CB=sqrt2. Sorry for that (CA=sqrt2 is correct).
To get EA then CE let EA=y and AB=1 (or any number but 1 makes the calculating easier)
Since triangle CED is a right triangle with a 45 degree angle, CE=ED=1+y. EB also =2+y
Since angle CEB is defined as a right angle and CBE (as shown by PreMath) is 30 degrees, CEB is a 30-60-90 triangle and CB will be twice CE or 2+2y
So by Pythagoras CE^2+EB^2=CB^2 or (1+y)^2+(2+y)^2=(2+2y)^2
Simplify and get 2y^2+2y-1=0 and so y=(sqrt3-1)/2 and EC=(sqrt3+1)/2
I hope that answers your question, Hussain. Now one can get CA=sqrt2 from Pythagoras. And looking at triangle CAD, the law of sines gives sinx/1=sin45/sqrt2 and sinx=1/2.
@@waheisel
That was beautiful math. It reminded me of a lot of stuff. Ty
I used to love math and physics with a dream to be get into cosmology. Due to personal reasons I'm now soon becoming a doctor.
It's good but I miss math and physics so badly 😅
@@yuda4626 Thanks for the kind comment Ty. When I was in school I also liked math, physics, and astronomy. I wasn't nearly good enough to make any of those a career. I also became a doctor. Even though you won't use so much math and physics in your career you can still enjoy your daily PreMath puzzle! Best Wishes
Question? I am very rusty in terms of geometry. The last course taken was during the 1984-1985 academic school year. From the diagram above might we say that 180 = 30 + ( x + 15 ) + Angle CAD and 180 = 45 + x + Angle CAD and we know by the rule of supplementary angles 180 = 45 + Angle CDB. So, by this rule, we know that Angle CDB has a value of 135. And, by the exterior angle rule given, 135 = x + Angle CAD. Based upon this information, drawn from the given, is it possible that angle x equals 25 and Angle CAD equals 110. Those numbers might be arising in my vision from the CBD smoke seeping in through the walls in this apartment unit; because, I do not smoke and do not physically tolerate it well.
A year or so after taking geometry, linear algebra taught that based upon the information given within a problems statement more than one "answer" might exist in the solution-space. And, that might be what is occurring here. If, I am not mistaken. So, we both might have settled on accurate solutions. I still a little puzzled by that external angle rule. Unless, Angle CAD can also have a possible measure of 105. Maybe, that is why it is labelled CAD?😄
nice one!
I like the ones when you need to make an extra lines in order to solve.
Before these things used to be a headache for me in class. I did it cause I was forced to and did the minimum.
Now it’s my entertainment and I can’t get enough of it. I rather watch this than a series.
Explained thoroughly well. Nice illustration
After watching few similar videos able to solve this one in mind. It's mostly in finding/drawing the correct isosceles, equilateral or congruent triangles with the equal length sides.
I admit I got stuck after step one. designating point P was a clever trick, but only worked out by chance, due to the particular properties of the two triangles. After it turned out that the vertex in D of APD is 60 degrees, the rest was easy!
No worries Philip👍 This was a challenging one indeed!
You are awesome. Keep persevering😀
Fantastic and clever solution! I couldn't made it that way!
I have another aprroach:
1. Draw the circle that contains point A,B,C. To do so, draw 2 perpendicular lines at the midpoind of AB and BC, these two lines meet at the center O. WE notice that AOB is the angle at the center and the angle ACB (= angle x +15 degrees) is the angle at the circumference.
2. Demonstrate that the angle at the center AOB is equal 90 degrees. Then, we can find the value of the angle ACB = 1/2 AOB= 45 degrees.
3. The angle x= angle ACB - 15 = 45-15= 30 degrees.
You cannot say that there is a circle which contains point A,B,C
@@emaceferli726
Yes, I can. Just label M and N the midpoint of AB and then AC. Draw 2 perpendicular lines to AB, and AC at these points which meet at the center O. Then we have OC=OA=OB, therefore we can draw a circle from the center O with the radius R=OC=OA=OB.
It is easy to see that the angle CBA=30 degrees (CMA is the exterior angle=45degrees).
The angle CBA is the angle at circumference of which COA is the angle at the center, so COA= 2 CBA=60 degrees----> the triangle COA is an equilateral one.
Let's say I is the midpoint of OA. The height from the vertex C to I is also the bisector of the angle ACO (=60 degrees).
Because the angle AMC=OMC=45 degrees so CM is the bisector of the angle OMA, so C,I,M are colleniar.
Thus the angle x= ACM= 30 degrees
Felt like a suspense story with a great payoff at the end! Well done good sir.
What if i use ratios to calculate x instead because the angles are propotional? The angles of
I got it. My sister told me the triangle CDP is an isosceles because angle PCD and PDC have the same 15 degree. Therefore, PC = PD. The triangle APD is equilateral, hence PC = PA. So angle PCA and PAC are 45 degree each. Consequently, angle ACD is 30 degree. That’s awesome!!! Thank you very much.
I had learned geometry 50 years ago. I have to confess this brings back a lot of memories.
Can you please let me know
How we concluded CP = PA?
Beautiful solution to the problem. Very much in the domain where math is art itself. I have opted for a cruder way (PS: I am an engineer). I dropped a perpendicular instead and used expressions for tan. Got two equations and soleved simultaneously to get the answer.
teacher I think you taught is very well, and made my ideas very clear, especially my observations.
Another flash of brilliance from the maestro, that was a superb demonstration, I'd never have got that, I like the way in many of your marvels that you add cunning lines that reveal the pathway, it's a joy to watch. Thanks again 👍🏻
You could've also just found angle BDC as it's a linear pair=135 and then as cdb is a triangle with angle sun property you can find angle b as 30 and then angle acd equals dbc so angle acd is 30
THE SHORT:
If you could draw DP so that DP=DB, on paper, you could also just as easily draw AP so that AP=AD=CP=PD, on paper... all anchored by the given angles of ADC & DCB.
THE LONG:
Imagine vertices ABC are 3 stars in a system, with D being a star exactly between star A and star B (let's say with Hubble, we're easily able to measure that the star D is exactly half way, in light years, between stars A & B, all three nicely aligned).
From ANY of these given stars, to any other star (in the ABCD set of star system), as well as, angles 15 & 45....nothing was given, could have been given, via assumption or "eye balling".... Everything came about through precise measurements via parallax + trigonometric, standard Candle method, etc.
Take one step back and ask yourself: --- what is the astronomical (if we took ABCD to be a set of 4 stars in the Milky Way) or geometric logic that said you could, for example, draw a precisely KNOWN line segment, like DP, from one vertex of BDC, to the opposite side, BC, in such a way that DP is exactly equal to one side, DB.... but NOT be able to do a similar thing, say, AP?
After all, AP=AD=DB, DB=DP, and AD=DB.... all anchored by the given angles of 15 (DCB) & 45 (ADC). For example, "P" can NOT be in any other place, along BC, such that AD is not equal to AP (due to the dictates of the angles 15 and 45, as given).
In other words, there's neither "wiggle room" for angles ADC & DCB, on the one hand, nor for point "P" along BC such that AP=AD=DB=DP is NOT true, on the other.
Man you're fascinating! Things like this never comes to the mind.
I did it in 2 or 3 steps by understanding the triangles are similar.
Is it correct approach?
I took Geo Trig last school year and felt very smart knowing how to solve everything. However when I saw this, it was not the case.
Thanks for explaining this very thoroughly. It was very fun and interesting to learn how to solve this Geo Challenge. Please make more!
My question is, when we create line DP = DB, how do you prove that point P were located on line CB?
Thanks a lot sir , i spent too much time thinking how to solve this and got to learn so much , indeed it was a wonderful experience.
I had no clue how to start the first step. Very nice and beautiful. Point p is strategical
Thanks for the excellent explanation. I stared at this problem for several minutes before coming up with the answer. It seemed obvious to me that angle X should be equal to angle ABC and ABC was obviously 30. based on the given angles, but while I could prove angle ABC couldn't figure out how to prove angle X.
ua-cam.com/video/Z67zK4B6cfU/v-deo.html😊😊
What software do you use to make these geometric shapes?
I used the Law of Sines and Cosines. It was simpler that way (for me anyway). Thanks.
Another method is to use sine theorem and a trig identity to prove sin x /cos x = (sqrt 3)/3 = tan x. So x =30. Less head scratching to figure out how to draw auxillary lines.
Thankyou Sir for the excellent and clear explanation. I succeeded to solve using the sin theoren (but it takes a lot of time).
😂
The area of triangle ACD is half of triangle ACB, it can also get the degree of X. Area = 0.5*length1*length2*sin(angle)
Hi, you could have separated triangle cdb and joined AD and DB by rotating clock wise triangle CDB. Making points A and B to be the same point. And you can get the answer in 2 steps. As rotating would create an isosceles triangle. Try it out
Rotating does not create an isosceles triangle. In fact, in the general case, the rotation is more likely to create a 4 sided polygon. In this particular case it creates a triangle, but not an isosceles one. To think this through, think about point C being split by the rotation, where C1 is the original point C, and C2 is created by the rotation. In this case, because angle ADC is 45 degrees and angle CDB is 135 degrees, after the rotation the line segments C1D and DC2 are collinear forming a straight line, with D in the center, since C1D and DC2 are the same length (but since they form a straight line they are not two sides of an isosceles triangle). So one side of the resulting triangle is C1C2, with the other two sides being AC1 and BC2 (A and B are the same point after the rotation). One of the angles in this new triangle is the 15 degrees from angle DCB prior to rotation, another angle is X degrees, formed from angle ACD prior to rotation (now AC1C2 after rotation). We know from the solution presented that X is not 15 degrees but is a value (trying not to spoil) such that none of the angles in the triangle formed by the rotation are the same.
i solved it using trigonometry
1. construction: draw ce perpendicular to be
2. angle cae=60(angle sum property)
3. angle bae=45-x
4. angle bea=45+x..........eq 1
5.tan 30=ae/ec=ae/eb+2db
6.tan 45=ae/eb+db
7. ae=eb+db
8. root3 ae=ae+db
9. (root 3-1)ae=db
10. eb=(2-root 3) ae
11. eb/ae=tan 75
12. angle bea=75..........eq 2
therefore from eq 1 and 2 x=30
Very thorough explanation. Saw similar videos and thanks for posting those. was wondering if there a pattern to solve such problems?
i guess in these kind of problems only the one who created them can solve them, or you can go through a long journey of trying all possible methods and theoremes to finally conclude it
Also you can use cos formula to calculate the angle
Благодарю за разбор интересной задачи.
Clearing away the cobwebs. 55+ years I was solving problems like this as part of a submarine fire-control problem (I.E. what angle to set the torpedo to intersect with the surface target) We had a mechanical analog computer, however, if it failed, you had to know how to do it by hand. Thank you. Narragansett Bay
I was just wondering how will you use that information, that those two parts of bottom line is equal, and I'm speechless how beautifully you created a triangle out of it and used that fact... I knew that equal sides of triangle subtend equal opposite angles in a triangle, but i wasn't smart enough to use that fact... Thank you sir, love from India..!
A class one teacher and a very challenging geometry problem.
You keep up the name Indians are great mathematicis
During my jee preparation (engineering entrance exam of India) i studied a theorem called m-n cot theorem, and really it gave the answer in just 5-10 seconds :) 😁
BTW great explanation sir😊
@Arnab Karmakar XI SC 6 Good job bro !!!All the very best 😀😀
I did it with some brute force trigonometry but it is easy to slip up on a figure when using calculator etc
Excellent! I learn geometry and English language in this channel! Thanks!
You're very welcome Sergio!
Glad to hear that!
Thank you for your nice feedback! Cheers!
You are awesome.😀 Keep it up!
Love and prayers from the USA!
As a first step use the 'Thales' theorem to get the 90 deg angle at P. Everything else follows automatically.
I used the sine rule twice and the sine addition angle formula, but this solution is far better.
Thank you for your feedback! Cheers!
You are awesome Barry.😀
Excellent tutorial. Uniques explanation. Enjoy watching and learning from your tutorial. Thanks.
I couldn't solve it thanks for this video 👍❤️
No worries Milli 👍
You are awesome. Keep persevering😀
will it work out if you simply express all sides with sine rule in a way that the sides cancel out and only an equation for sin(x) remains, leaving an answer of sin(x)=1/2, which due to that 45 degree angle makes 30 deg the only solution?
I think the question may be solved by applying basic triangle rules. I tried by following method-
1. By external angle theorem: angle DBC = 30°
2. In triangle DBC, if length of DB is 'a' (against 15°), then CD is 2a (against 30°)
3. Now, in triangle CAD, AD = a, CD = 2a
4. Angle ACD = x, Angle ADC = 45° = y, Angle CAD = z
5. Angle z = 2 (Angle x) [because CD = 2 (AD)].
6. As, Angle x + y + z = 180°
x + 45 + 2(x) = 180°
3x = 180° - 45° = 135°
x = 135°/3
x = 45°
Cool solution
But x = 30°
angle double doesnt mean side is double...its a sine relation
There is no rule in a triangle as you wrote "if length of DB is 'a' (against 15°), then CD is 2a (against 30°)". By this info you can just write: CD>AB.
This question is a lot easier when you substitute the bottom two same sides to be 1 each, then use sine rule and cosine rule to find x
Is it only me who thought that if AD=DB, CD is the median line of the triangle in C, thus meaning in cuts the angle in half? And if DCB is 15, ACB would be 30?🤔🤔
You’re correct especially if you check with Sine law
Actually, that is not always the case, as shown in this problem here. The median of a triangle in a given vertex is not always equal to the bisector of the relative angle. They're the same in some cases, for example if the triangle is equilateral, it is true for alla the vertices.
@TJ the sine law is applied within a given triangle, in this case you can say tha
sin(15°)/(DB) = sin(30°)/DC
And
sin(x)/AD=sin(135-x)/DC
You're mistaken between median line and angle bisector.
This is proven false in this example so if it cit ACB in middle = ACD = DCB =15° which is not the case
Solved using a very long method. A shorter way to do it is to extend CD to P so that CD=DP. Triangle DBP can be proved congruent to triangle CAD and angle X will be equal to angle DPB. DPB will be first calculated to be 30 degrees.
I solved it using sine rule, without any construction. Good problem !
How pls tell me
Great Subramaniam dear.
Thank you for your nice feedback! Cheers!
You are awesome.
Keep rocking😀
I used the same method as well! Law of sines to get it
Very nice way. Before showing your solution, I got to the trigonometric expression x=45°-arctan(2-sqrt(3)) at the end of my way...
Me too. Then, as tan 15°=2 - √3 (use the compound angle formula to evaluate tan 15°=tan(45°-30°) to get this), we get x=45°-15°=30°.
I didn't spot the geometric solution until I saw the construction of AP in the video.
Glad to hear that!
Thank you for your nice feedback! Cheers!
You are awesome Dreael.😀
@@PreMath And here the details for all others: Draw the height from point C to the A-B line. Call the foot point as E. In my way, I also calculated the angle at point B first, i.e. 30°. The triangle BCE shows a 30°/60°/90° triangle. So when BC has a length of 2 units, then CE has lenght 1 unit and BE is sqrt(3) units long. CD then is sqrt(2) units long because of the 45° angle and DE is 1 unit long (40°/45°/90° triangle CDE). Lengt BD = BE - DE = sqrt(3) - 1. Because AD = BD =sqrt(3)-1 units then AE = DE - AD = 1 - (sqrt(3) - 1) = 1 - sqrt(3) + 1 = 2 - sqrt(3). Focus on the right triangle ACE where the angle ACE can be calculated as tan(ACE) = AE/CD = (2 - sqrt(3))/1 = 2 - sqrt(3). And important to know: tan(15°)=2-sqrt(3) is also a special value (can be easily proved on a 75°/75°/30° isosceles triangle). The actual result x is just the difference of the angle BCE (45°) and ACE (15°) in this case.
Gosto muito de seus ensinamentos, obrigada!
De nada, Sonia!
Obrigado! Saúde!
Continue agitando😀
Amor e orações dos EUA!
Very clever problem. I thought you would need trig so I gave up on it. Very impressive that you can solve this without needing to use trig.
There are many ways to solve this kind of problem!
Thank you for your nice feedback! Cheers!
You are awesome Lance😀
could use trig by drawing a perpendicular line from C to P to form right triangle CPA. Triangle CPB is a 30-60-90 right triangle. label the sides as 1, sqrt 3 (or 1.732) and 2 let's label angle CPA as 'y' . y + x =45 degrees and you now have and 45-90-45 with sides 1, 1, and sqrt 2
degree right-triangle line PD =1 and CD= sqrt 2 . Using ASA (15 degree, sqrt 2 and 135 degree) for triangle CDB gives 0.732 for DB , but DB = AD thus PA = 1-0.732 = 0.268. Since the right triangle CPA has the sides 1, 0.268 then using SAS (1, 0.268, and 90 degree) gives 15 degrees for angle y,
but y+x=45 degree hence x=30 degree
Trigonometry makes the whole thing much easier. Actual challenge is to solve it without trigonometry. ☺️
30 degrees ( since ACB angle is 45 degrees).
Angle ABC is 30 degrees, draw the line DE parallel to CB ( which crosses AC in E). Length of DE is 1/2 of CB, since AD = DB. Then draw two lines from D and E, each perpendicular to CB.
Denote that the only thing which needed is to know that in the right triangle with angles 30/60, the side opposite to 30 degree angle is 1/2 of side opposite to 90 degree angle, and side opposite to 60 degree angle is √3/2 of it. For that no nowledge of trigonometry is required, only Pythagorean theorem.
It was an easy puzzle! You could've solved it with a much easy technique. ∠CDA + ∠CDB=180°[Linear Pair] . So, ∠CDA is given as 45°. Then, 45°+ ∠CDB=180°. ∠CDB=135°. ∠B+∠CDA+∠DCB=180°[Angle sum property of a triangle]. Then if we construct a similar congruent triangle to CAD say CAE. we get x+x+15°+45°+45°=180°[angle sum property of a triangle]. we get the value of x as 45°. So, this was a more easy technique😀😀
Easy approach
That's the one I was thinking of aswell.
But the value of x is 30°, it says.
03:15
Honesty i couldn't solve it but very good question to practice thanks😊
No worries Pranav dear 👍
You are awesome. Keep persevering😀
@@PreMath thanku
I haven't played with this stuff for 60 years but that was beautiful, man. Kudos and thank you.
Just sane... :^) Saint
So difficult
This problem gave me a lot to think about
No worries 👍
You are awesome. Keep persevering😀
I used trigonometry to solve this but this method is much simpler. Nice solution
I have failed to solve it.🙁
No worries 👍 This was a challenging one indeed!
You are awesome Mustafiz. Keep persevering😀
I remember a short cut, but I'm old and don't know where I learned it. Basically if the line segments are the same then the angle x = the angle CBD. so all you have to do is solver for CBD. Angle ADC may have to be 45 though(it might have to do with trig). I don't remember.
I used sin rule. Sin15/DB = sin30/CD. I get DB = 0.518CD. and then sinx/DB = sin(135-x)/CD. I substituted DB = 0.518CD into the second sin rule equation. CD got cancelled out and after using some trigo identities, I managed to solve for x as the answer given. Good qns!
Great Garrick.
Thank you for your nice feedback! Cheers!
You are awesome.
Keep rocking😀
Following this method - if you stick to using surds and expand the sin(A-B) you can rearrange and get tan(x) = 1/Sqrt(3) and this x is 30 degrees.
Yes that’s right. I was lazy to deal with surds so I used decimals. Same answer will be derived :)
@@starpupil1843 AD = DB. so it’s the same. My aim was to form two equations involving DB and CD.
@@starpupil1843 u r welcome! My weakness involves me not knowing what line to draw or what to substitute (in calculus) to get to my solution faster. I deal with what I see and what I can find base on what I see. Hahs
Extend line CA and obtain exterior angle (180-x-15) =x+ ang CAD thats one equation. Another eqn is using triang.CAD :
x + Ang CAD =135. We have two eqns and two unknowns which we can solve to get x=30
Excellent explanation . Thank you .
You can use m-n cot theorem to solve it in 3 steps:
2.cot(135°) = cot(x) - cot(15°)
cot(x) = cot(15°) - 2 = 2+√3 -2 = √3
X = 30°
This question is actually very easy
Draw a perpendicular to line AB, let the point be H
Assume length AD as l
Write Tan45 and tan30.
Solve for AH and CH in terms of l
Find CH/AH, this is tan(45-x)
We get tan(2-√3) = 45-x
Sir if we use in triangle ACD tan 45° so AD=AC Therefore triangle ACD isosceles.so,
Thank you, very well explained, i appreciate the attention to detail in the explanation.
I solve it by using sin cos tan. Your way is much simpler and clearer