An alternative (purely geometric) solution could look as follows. As angle ADB is equal to 45° and angle ACB is equal to 30°, the angle CAD is equal to 15°. If one joins point D with a point E on AC such that angle ADE is equal to 15°, then |AE| = |ED| = |DC| = |BD| and angle DEC is equal to 30°, which implies that angle BDE is equal to 60° and due to |DB| = |DE| that triangle BDE is equilateral and that the angle BEC is equal to 90°. As |AE| = |EB| and the angle BEA is equal to 90° (180° - [angle BEC]), the triangle BEC is a right and isosceles triangle and, as a consequence, the angle BAC is equal to 45°, which implies that the angle BAD is equal to 30°.
I must admit I shook my head when I saw the thumbnail - not enough information! The equal sides are not given in the diagram. Anyway, I let y = AD instead,
You could get sin(15) from sin(30) = 1/2, if not with a half angle formula then with a system of equations coming from the double angle formula and the pythagorean identity
Yes, the trick is that there are two equal sides, the second being AD, and that's enough for sides to cancel out when applying the sine law on the two triangles. Here is my solution: - using the sine law BD/sin 15 = AD/sin 30 and BD/sin θ = AD / sin (135-θ) - dividing the corresponding sides and rearranging gives a single equation, sin(135-θ) / sin θ = sin 30 / sin 15 = 2*cos 15 - the difference law gives √2/2 (sin θ + cos θ) / sin θ = 2*cos 15 - finally, expanding the left side gives cot θ = √2*2*cos 15 - 1 = √2*2*cos (45-30) - 1 = √2*2*√2/2*(√3/2 + 1/2) - 1 = √3 + 1 - 1 = √3.
Draw a circle around D with radius |DC|. Let the intersection of the circle with AC be E. According to Thales' theorem, triangle BCE has the angles 30, 90 and 60. Since |BD| = |DE|, triangle BDE is equilateral. Since angle DAE and angle EDA are 15 degrees, |AE| = |ED| = |EB| and therefore triangle BEA is isosceles. Therefore angle theta is 45-15 = 30 degrees (base angles triangle BEA are (180-90)/2 = 45 degrees).
@@activatewindows7415 This solution doesn't require accurate drawing. It is the most elegant solution, because it doesn't need any trigonometric identities - just Thales theorem and elementary geometry.
Michael if you want to improve your geometry you need to give up analytic solutions don’t fall into the trap I fell into,solve them geometrically it’s tricky but it’s worth it because you gain better insights of finding key points which leads to more elegant and less messy solutions..Another issue with analytic solutions is that when solving more complex geo problems you often stumble upon 4th and higher degree polynomials which are usually not possible to solve,for example try the generalized ladder and square problem from mind your decisions analytically-completely impossible.lastly geometric solution are just more satisfying as they are require more creativity and less brute force.
When presented with a geometry problem, turning it into algebra is like taking a long and complicated detour to get to the destination, rather than taking the freeway. It is really much more elegant and satisfying to solve this problem using theorems from geometry.
without drawing anything extra just use law of sines, u get two equations and u solve for the theta variable only a/sin(A) = b/sin(B), A = theta + 15degrees, B = 180degrees - 45degrees - theta, sin(B) further simplified to sin(theta + 45degrees) by removing the negatives so u have 2 equations a/sin(theta + 15degrees) = b/sin(theta +45degrees) b/sin(135degrees) = 0.5a/sin(15degrees) substitute a or b, and the other will also be eliminated, leaving theta to be solved for no calculators needed, just the table for 30 45 60 degrees and the angle sum formula u need to find the ratio of sin to cos and then solve for tan from the table, giving 30 degrees as the answer (need to simplify and rationalize all the square roots to get the table value)
The logical way to start is ususlly to fill in complimentary angles and the angles that directly follow from sum of angles in a triangle is 180 degrees.
This is mental arithmetic only. Within seconds, I had the answer. Watching you, I had my head in my hands, saying "no, no, no". This is exactly how to complicate a very simple "problem".
That is a very classic geometry problem. Here is one alternative solution in pure geometric construction. Suppose point P be the symmetric point of D about the line AC, we get a regular ΔCPD and BD=CD=PD. Note that ∠DBP = ∠DPB = ∠CDP/2 = 30⁰, and ∠DAP = 2∠CAD = 30⁰ = ∠DBP, it follows that A/B/D/P are co-cyclic. Therefore, the θ = ∠BAD = ∠BPD = 30⁰
Beautiful solution! But some obscurities. In particular, why is angle DPB = CPD/2? I got triangle CBP is equilateral. Then from DC=DB=DP, P is on the semicircle with D as centre and CB as diameter, so angle CPB is a right-angle; whence angle DBP = angle DPB = 90 - 60 = 30 Or (more in the spirit of your thinking), angle DBP = angle DPB (isosceles triangle) = 1/2 x angle CPD = 30 (angle at centre of circ on CPD vs angle at circumference)
Neat problem. 🙂 The only minor thing I noticed in the video is, the way he wrote and drew everything, he assumed point E fell outside the triangle, but at least in principle going into the problem at first you don’t actually know if E is outside the triangle or if it lies on the line segment BD. I don’t think this is a big deal though, it just means that y in the first part of the proof might be negative as well as the corresponding small angle. That doesn’t seem to change any of the algebra though, and in fact as shown in the video when you solve for x/y you get a positive ratio instead of a negative one which means y is in fact positive and therefore outside the triangle.
He does say that angle ABD is not 90°, so if you can assume the angle of 45 and lengths are correct, theta can't be 45° also then. Therefore, to extend theta to 45°, you would need E to be outside of the original triangle.
I have seen problems like these where a rough sketch makes it look like 90 degrees when in fact it can be 88 to 92 degrees excluding 90 degrees. It is a way to challenge a drawn solution over an analytic solution based, as you observe, on some grounded foundation principles
@@monzurrahman8307 It is a good point for sure. Is the angle obtuse or acute? I suppose an alternative approach is to use differential calculus on an assumption that a line drawn from BDC passing through point B will be at its shortest length when it is perpendicular to the line. Think describing an arc with compass point centered on point B. But then we have to factor in that point D bisects line BC and this actually anchors point E as an extension to line BC. Try drawing it with ruler, compass and protractor either on screen or on paper (I think paper is the best for hands on) Share your findings here?
Honestly: Without loss of generality put midpoint of equal lengths at origin, line along x axis, equal lengths = 1. Solve for intersection of lines y = -(1/sqrt(3))(x-1) y = -(x+1) No pythagoras, no dropped perpendiculars, no tan angle addition identities
I knew that problem and I found following pure geometric solution: The first steps are the same as in the video, until AE = x + y. The triangle AEC is half an equilateral triangle, the angles are 30,60,90. -> AC = 2x + 2y. Construct point F on line BCDE, on the left side of E, with EF = EB = y. Look at triangle ACF: it is isosceles because of AC = CF = 2x + 2y. The angle on the top is 30, so the angles on the bottom must be 75. Angle AFE = 75 Angle ABE = 75 because triangles AEF and AEB are congruent. Angle DBA = 105 and angle BAD = 30.
The worst problems like this usually include a sketch that looks to be a right angle but may be 87 to 93 degrees or even worse: 88 to 92 degrees Just a suggestion going forward: is it reasonable to do several (linked?) videos on this problem and solution solving? My reasoning is that viewers and student may (will?) find it helpful to see several ways to solve this problem. I hope that is a reasonable way for learners to find their preferred working methods aware that several options do exist bringing awareness of wider solution solving in general
Thats like solving a quadratic equation with calculus. If this is a geometric problem, then it should be solved in ageometric way. Second, how do you know apriori that point e is not in cd? (You have to prove this part, since drawings are sometimes misleading.)
the fuck u mean point e is in cd because he made it not be in cd, u dont need to prove a fucking placeholder variable, e is that point specifically set outside i can let f=10 and i dont need to prove that because i made it up to solve things ffs, f in this case is a placeholder the same as e in this video and unless u r blind and have no common sense, how the fuck can e be in cd? they are like 2 light years apart, this is literally a visual problem, not some quantum bullshit
BRO U DONT NEED TO BE A GENIUS TO KNOW E IS NOT IN CD THEY ARE LITERALLY 2 LIGHT YEARS APART AND IF U MEAN BD, THEN U DONT NEED TO KNOW, BECAUSE IT DONT EVEN MATTER, THE PROCEDURE DOESNT EVEN CHANGE
I solved the problem in a totally different way, using the sin and cos laws to complete each triangle. It definitely felt like a less clever solution but it was super satisfying and a really great way to review a bunch of elementary trig. Definitely keeping this problem in my back pocket for students!!
I did the same construction of E, and used an angle sum formula, but more simply for both parts. As no lengths or areas are required, without loss of generality, define AE=1. By well known ratios for 30°-60°-90° and 45°-45°-90° triangles, EC=√3 and ED=1. DC=EC-ED=√3-1, but BD=DC, so BD=√3-1. EB=ED-BD=1-(√3-1)=2-√3. I'm sort of synced up here with Dr Penn at 6:15, same ratios expressed differently. Defining φ=∠EAB, tanφ=(2-√3)/1.=2-√3. I suspected φ=15°, and a quick check on a calculator indicated this was worth pursuing. Rather than use a half angle formula and have to untangle nested square roots, I opted to work up and calculated tan(2φ)=2tanφ/(1+tan²φ) = 2(2-√3)/(1+(2-√3)²) = (4-2√3)/(1+4+4√3+3) = (4--2√3)/(8-4√3) = (4--2√3)/2(4--2√3) = ½. So 2φ is either 30° or 210°, and only 30° makes sense for the diagram, so indeed φ=15°. Clearly φ+θ=45° from ⊿ADE, so θ=30°. And IMHO, that's a good place to stop.
Impressed by the pure geometry solutions posted. Like several others, I was drawn to the sine rule approach. Using it on triangle ACD to show that |AD|=|AC|/Sqrt[2] and then applying it to triangles ABD and ABC and eliminating sine of angle ABD leaves sin(theta)=sin(theta+15)/Sqrt[2]. Rather than hammering this out with addition formulae and tan or cot, in this case (particularly recalling sin(45) =1/Sqrt[2]), theta is 30 degrees by inspection.
Much easier by using law of sines. For the larger triangle, Sin (a+15)/2x = sin 30/z and using the smaller triangle, sin 45/z = sin a/x. Simplfying, we get inverse cot of 1.732 which is 30 degrees. ATAGPTS
Can we solve this by drawing a parallel line to BC passing through point A and then using supplementary and complementary angle rules for parallel lines..I am getting stuck though idk
@@rarhs Cleanest shortest geometric solution posted by @papafragen above - Uses 1 circle construction followed by Thales theorem and isoceles theorems...done.
My only marginally different solution was to set the x in the video to 1, without loss of generality, which then immediately gave me tan 60° = sqrt(3) = (y + 2)/(y + 1), and therefor y = (sqrt(3) - 1) / 2. This then implied that atan(angle EAB) = y / (y + 1) = 2 - sqrt(3), and therefor angle EAB = 15°, and therefor theta = 45° - 15° = 30°.
what i did was to find the angle EAB, which is arctan(y/(x+y)) which comes out to 15°, and then just subtract that from 45°; by that method you don't have to rationalise denominators etc anywhere
From the law of sines, x being the unknown, sin(x+15)=sin(x)/sin(45) => Dividing both sides by cos(x), the equation becomes tan(x)/sin(45)=cos(15).tan(x)+sin(15) => tan(x)=sin(15).sin(45)/(1-cos(15)sin(45)) Finally, tan(x)=1/sqrt(3) and x=30
This problem can be solved without any construction. As we see that AD is a median of triangle ABC it divides the triangle into two with equal areas. Since angle CAD is 15, using sine formulas for area of the triangles in two different ways and remembering that angle ABC is (180 - (theta + 45)) we get four equations for the area of triangles. Which can be solved very easily for tan theta. Without so much of mathematical jugglery.
Gentile e perspicuo Michael, se posso permettermi, ti dico: assegnato un valore unitario a BD=DC, applicando il teorema dei seni prima al triangolo ACD per ricavare AD e poi il teorema di Al-Kashi (Carnot) al triangolo ADB, per ricavare AB, applicato nuovamente il teorema dei seni al triangolo ABD, l'angolo theta si ricava molto più facilmente della tua pur pregevole prolusione. Un caro saluto.
I just started on the problem and then asked AI what the answer is. She told me in about 3 seconds so that is how I am going to work the problem. I will never have to do this problem in my everyday life and I really need to mow the yard and do some other things so I really like asking AI. After all, isn’t that why we even have computers in the first place to make our lives easier. I need to figure some income tax info too so I think I will use TurboTax and then I will have time to also do the grass edging after I mow.
Drop a perpendicular BH to AC and observe that triangle BHD is equilateral. Notice that angle ADH is 15 degrees, so that triange ADH is isosceles and AH = DH=BD. Hence, points A, B, D lie on the circle centered at H. The inscribed angle BAD then is equal to one half of the 60 degree central angle BHD, that is, 30 degrees.
DAC = 180° - DBA = 180° - 45° = 135° ACD = 180° - DAC - CDA = 180° - 135° - 30° = 15° ACB = 180° - CBA - BAC = 180° - 30° - BAC = 150° - BAC ACB = ADB + ACD = ADB + 15° BAD = 180° - DBA - ADB = 180° - 45° - ADB = 135° - ADB Three Equations with using BAD = BAC: (1) ACB + BAC = 150° (2) ACB - ADB = 15° (3) BAC + ADB = 135° (1) - (2) BAC + ADB = 135° I just realized that D should be exactly in the middle of AB when I let the video play for 5 seconds :D I was just wondering why my system of equations is underdetermined.
The pure geometry method is much more satisfying... and if you are lucky enough to construct the perpendicular from B to AC (call the new point E) and then join ED... it all falls nicely into place. Mind you, I did flail around for a while before trying that.
After reading some of the comments, I’m thinking pure luck I got the 30 degree answer. Geez I’m glad I will never have to be in a geometry class or a calculus class ever again. And my grandkids can go online for help.
Since bdc is a straight line,angle abd is equal to 180-(45+30)=105 degrees So angle bad is equal to 180-(105+45)=30 degrees It is a very simple problem.
Aight... Math time. The triangle ADC can easily give the three angles as 135 + 30 + 15, and the triangle ABD has angles theta, 45, and 180-45-theta. Now let's use the law of sines. For any triangle with corners ABC, sin(A)/BC = sin(B) / AC = sin(C) / AB Since AD is shared between the two triangles, this simplifies to these two equations. Let y = AD and x = BD = DC sin(theta) / x = sin(180 - 45 - theta) / y sin(15) / x = sin(30) / y since sin(180-theta) = sin(theta): x = y*sin(theta) / sin(45 - theta) x = y*sin(15) / sin(30) Combine both, and... y*sin(theta) / sin(45 - theta) = y*sin(15) / sin(30) sin(theta) / sin(45 - theta) = sin(15) / sin(30) theta = 15
A Simple way is to draw a height line BE, the height is the same as BD and DE (because BEC on the same circle, C is the center). So angle EAD equals angle EDA. So DE=EA=EB. So angle CAB=45 so angle dab=30
My math teacher in high-school asked us how does on solve math problems? But nobody could answer including the teacher. Many years later, helping old people to use their phones and young how to do math, I realised the solution. The secret using phones are the same as solving math problems. It is very simple: Do something regardles what. ( don't just sit there and look at it ) . It sounds silly but when you do something it may be wrong or not help but that only means do something else regardless what. This way you can solve all problems. I am now 82 year old retired technical director, I tell you this is true.
that bottom left angle is 90+15 degrees = 105 degrees. Total degrees are 180, so (using the 45 degree and 30 degree ratio) the top angle must be 45 degrees total and the unknown angle must be 2/3 of the 45 degrees or 30 degrees. It's been a while since I delved into geometry so I've forgotten much of the nomenclature, but not the process. Trigonometry is for losers....(kidding). 3x2(9yz)4a
This video got me to thinking about other contructable whole numbered angles. I tried google but couldn't find any site that has a complete list of the possible angles that can be constructed. (1°~180°)
Faster way to solve: B=135-θ (angle sum of triangle) ∠DAC=15 (angle sum of triangle) Set BD=CD=1 unit AD=0.5csc15 (sine law) 2sin(135-θ)sin15=sinθ (sine law) tanθ=(√(2)sin15)/(1-√(2)sin15) tanθ=1/√3 θ=30
Let T = theta. AD is common side of 2 triangles and BD = DC, so *sine rule* can solve this problem in the simplest way. Angle B = 180 - 45 - T = 135 - T and Angle DAC = 45 - 30 = 15. Therefore, using *sine rule* sin(135-T) / sin T = sin(30) / sin(15) = (1/2) / [(√3-1) / 2√2] = (√3+1)/√2 sin(135-T) = sin(45+T) = (sin T + cos T)/√2 = sin T (1 + cot T)/√2 Therefore, (√3+1)/√2 = sin(135-T) / sin T = (1 + cot T)/√2, so cot T = √3, so *theta = T = 30 deg*
Ok. I’m no math genius. And by the time you were done, I was lost and confused. But I realized that the angle was thirty degrees. And I did that using the calculator on my iPhone. And that’s what you ended up with, while making me feel like how I felt in math classes through primary school to college. Blank stare on my face and huh? I mean it took me less than a minute.
I found a shorter solution. If x = BD = DC, and y = AB, By the Law of Sines in triangles ABD and ABC, sin θ/x = sin 45°/y and sin (θ+15°)/(2x) = sin 30°/y. Dividing both members, we can eliminate easily x and y, and then sin (θ+15°) sin 45° = sin θ. This equation has the obvious solution 30° because sin 30° = 1/2 = (1/√2)² = sin² 45°. You can solve the last equation doing some tricks I could post.
I never recall the exact values of sin 15° and cos 15°, so I did the following. Recall that sin (θ+15°) sin 45° = sin θ, thus -sin (θ+15°) sin 45° = -2sin θ. Then, cos (θ+60°)-cos(θ-30°) = - 2 sin θ. Expanding the left side, etc. we find 1/2 (3-√3) sin θ = 1/2 (√3-1) cos θ. Simplifying, tan θ = 1/√3, so θ = atan (1/√3) = 30°.
Way too complicated! This was a 20 second problem. We can apply numerical values to the sides since we have 30˚. AC = 4, AE = 2, BD = DC = √3 -x. ED = x+√3=AE = 2, EB=2x, EAB = arc tan 2-√3 = 15˚, since EAB = DAC = 15˚, therefore BAD = 30˚.
Maybe I just got lucky but the bottom straight line =180deg subtract 45 and 30 deg =105 so ABD is 105 add the 45 deg to that then subtract that from 180 and I get 30 deg
By inspection, angle CDA =135 degrees, so angle DAC =15 degrees. Applying the law of sines yields the dimensions AD & AC with respect to X. Then apply the law of cosine to triangle ABC (with the obtained value of AC, BC = 2X and the angle between them = 30 degrees) to find AB (alternatively: apply the law of cosines to ABD with the obtained value of AD, BD=X and angle BDA=45 degrees). Now apply the law of sines: X/ sin(theta) = AB/ sin(45).
An alternative (purely geometric) solution could look as follows. As angle ADB is equal to 45° and angle ACB is equal to 30°, the angle CAD is equal to 15°. If one joins point D with a point E on AC such that angle ADE is equal to 15°, then |AE| = |ED| = |DC| = |BD| and angle DEC is equal to 30°, which implies that angle BDE is equal to 60° and due to |DB| = |DE| that triangle BDE is equilateral and that the angle BEC is equal to 90°. As |AE| = |EB| and the angle BEA is equal to 90° (180° - [angle BEC]), the triangle BEC is a right and isosceles triangle and, as a consequence, the angle BAC is equal to 45°, which implies that the angle BAD is equal to 30°.
I was posting the same solution
I use to solve geometric problems with geometry
If I'm not mistaken right at the end you should be referring to triangle BEA as the right isosceles triangle. Nice work nonetheless🙌
purely geometric, always a more elegant solution in my book.
Definity not as easy or as simple as the trigonometric way.
good
I must admit I shook my head when I saw the thumbnail - not enough information! The equal sides are not given in the diagram. Anyway, I let y = AD instead,
You could get sin(15) from sin(30) = 1/2, if not with a half angle formula then with a system of equations coming from the double angle formula and the pythagorean identity
wym exact value
it can be found from the trig table for 30 45 60 degrees
by using the sum formula for trig, sin(45-30)
That's the mathematical version of clickbait
Yes, the trick is that there are two equal sides, the second being AD, and that's enough for sides to cancel out when applying the sine law on the two triangles. Here is my solution:
- using the sine law BD/sin 15 = AD/sin 30 and BD/sin θ = AD / sin (135-θ)
- dividing the corresponding sides and rearranging gives a single equation, sin(135-θ) / sin θ = sin 30 / sin 15 = 2*cos 15
- the difference law gives √2/2 (sin θ + cos θ) / sin θ = 2*cos 15
- finally, expanding the left side gives cot θ = √2*2*cos 15 - 1 = √2*2*cos (45-30) - 1 = √2*2*√2/2*(√3/2 + 1/2) - 1 = √3 + 1 - 1 = √3.
@@bonzinip That is pretty much exactly what I did! Well done.
No Good Place To Stop😢😢😢
I really felt the absence of this.
Draw a circle around D with radius |DC|. Let the intersection of the circle with AC be E.
According to Thales' theorem, triangle BCE has the angles 30, 90 and 60.
Since |BD| = |DE|, triangle BDE is equilateral.
Since angle DAE and angle EDA are 15 degrees, |AE| = |ED| = |EB| and therefore triangle BEA is isosceles.
Therefore angle theta is 45-15 = 30 degrees (base angles triangle BEA are (180-90)/2 = 45 degrees).
Very elegant.
@@activatewindows7415 This solution doesn't require accurate drawing. It is the most elegant solution, because it doesn't need any trigonometric identities - just Thales theorem and elementary geometry.
Very nice!
Michael if you want to improve your geometry you need to give up analytic solutions don’t fall into the trap I fell into,solve them geometrically it’s tricky but it’s worth it because you gain better insights of finding key points which leads to more elegant and less messy solutions..Another issue with analytic solutions is that when solving more complex geo problems you often stumble upon 4th and higher degree polynomials which are usually not possible to solve,for example try the generalized ladder and square problem from mind your decisions analytically-completely impossible.lastly geometric solution are just more satisfying as they are require more creativity and less brute force.
When presented with a geometry problem, turning it into algebra is like taking a long and complicated detour to get to the destination, rather than taking the freeway. It is really much more elegant and satisfying to solve this problem using theorems from geometry.
I guess that wasn't a good place to stop.
So tan(75°)=2+sqrt(3)
without drawing anything extra
just use law of sines, u get two equations and u solve for the theta variable only
a/sin(A) = b/sin(B), A = theta + 15degrees, B = 180degrees - 45degrees - theta, sin(B) further simplified to sin(theta + 45degrees) by removing the negatives
so u have 2 equations
a/sin(theta + 15degrees) = b/sin(theta +45degrees)
b/sin(135degrees) = 0.5a/sin(15degrees)
substitute a or b, and the other will also be eliminated, leaving theta to be solved for
no calculators needed, just the table for 30 45 60 degrees and the angle sum formula
u need to find the ratio of sin to cos and then solve for tan from the table, giving 30 degrees as the answer (need to simplify and rationalize all the square roots to get the table value)
This would've been my approach as well. Felt like it was made unnecessarily complicated in the video.
The logical way to start is ususlly to fill in complimentary angles and the angles that directly follow from sum of angles in a triangle is 180 degrees.
I used law of sines
Same. It was a painful derivation involving completing the squares and cosine of 15 degrees.
the way u wrote the bracket for the last tangent equation there was wrong
This is mental arithmetic only. Within seconds, I had the answer. Watching you, I had my head in my hands, saying "no, no, no". This is exactly how to complicate a very simple "problem".
That is a very classic geometry problem. Here is one alternative solution in pure geometric construction. Suppose point P be the symmetric point of D about the line AC, we get a regular ΔCPD and BD=CD=PD. Note that ∠DBP = ∠DPB = ∠CDP/2 = 30⁰, and ∠DAP = 2∠CAD = 30⁰ = ∠DBP, it follows that A/B/D/P are co-cyclic. Therefore, the θ = ∠BAD = ∠BPD = 30⁰
Beautiful solution! But some obscurities. In particular, why is angle DPB = CPD/2?
I got triangle CBP is equilateral. Then from DC=DB=DP, P is on the semicircle with D as centre and CB as diameter, so angle CPB is a right-angle; whence angle DBP = angle DPB = 90 - 60 = 30
Or (more in the spirit of your thinking), angle DBP = angle DPB (isosceles triangle) = 1/2 x angle CPD = 30 (angle at centre of circ on CPD vs angle at circumference)
Neat problem. 🙂 The only minor thing I noticed in the video is, the way he wrote and drew everything, he assumed point E fell outside the triangle, but at least in principle going into the problem at first you don’t actually know if E is outside the triangle or if it lies on the line segment BD.
I don’t think this is a big deal though, it just means that y in the first part of the proof might be negative as well as the corresponding small angle. That doesn’t seem to change any of the algebra though, and in fact as shown in the video when you solve for x/y you get a positive ratio instead of a negative one which means y is in fact positive and therefore outside the triangle.
He does say that angle ABD is not 90°, so if you can assume the angle of 45 and lengths are correct, theta can't be 45° also then. Therefore, to extend theta to 45°, you would need E to be outside of the original triangle.
I believe it doesn’t really matter. If it were to fall inside the triangle everything would hold but y would be negative
I have seen problems like these where a rough sketch makes it look like 90 degrees when in fact it can be 88 to 92 degrees excluding 90 degrees. It is a way to challenge a drawn solution over an analytic solution based, as you observe, on some grounded foundation principles
@monzurrahman8307 if you extend the angle then of course it is correct.
But how do you know that theta is not bigger than 45?
@@monzurrahman8307 It is a good point for sure. Is the angle obtuse or acute? I suppose an alternative approach is to use differential calculus on an assumption that a line drawn from BDC passing through point B will be at its shortest length when it is perpendicular to the line.
Think describing an arc with compass point centered on point B.
But then we have to factor in that point D bisects line BC and this actually anchors point E as an extension to line BC.
Try drawing it with ruler, compass and protractor either on screen or on paper (I think paper is the best for hands on)
Share your findings here?
I'm still going...
Honestly:
Without loss of generality put midpoint of equal lengths at origin, line along x axis, equal lengths = 1. Solve for intersection of lines
y = -(1/sqrt(3))(x-1)
y = -(x+1)
No pythagoras, no dropped perpendiculars, no tan angle addition identities
I knew that problem and I found following pure geometric solution: The first steps are the same as in the video, until AE = x + y.
The triangle AEC is half an equilateral triangle, the angles are 30,60,90. -> AC = 2x + 2y.
Construct point F on line BCDE, on the left side of E, with EF = EB = y.
Look at triangle ACF: it is isosceles because of AC = CF = 2x + 2y. The angle on the top is 30, so the angles on the bottom must be 75.
Angle AFE = 75
Angle ABE = 75 because triangles AEF and AEB are congruent.
Angle DBA = 105 and angle BAD = 30.
Excellent tutorial. I learn something from your approach to this specific problem.
The worst problems like this usually include a sketch that looks to be a right angle but may be 87 to 93 degrees or even worse: 88 to 92 degrees
Just a suggestion going forward: is it reasonable to do several (linked?) videos on this problem and solution solving?
My reasoning is that viewers and student may (will?) find it helpful to see several ways to solve this problem.
I hope that is a reasonable way for learners to find their preferred working methods aware that several options do exist bringing awareness of wider solution solving in general
Thats like solving a quadratic equation with calculus.
If this is a geometric problem, then it should be solved in ageometric way.
Second, how do you know apriori that point e is not in cd? (You have to prove this part, since drawings are sometimes misleading.)
I also had this habit of solving geometry problems with trigonometry,I missed a lot of insight because of this
the fuck u mean point e is in cd
because he made it not be in cd, u dont need to prove a fucking placeholder variable, e is that point specifically set outside
i can let f=10 and i dont need to prove that because i made it up to solve things ffs, f in this case is a placeholder the same as e in this video
and unless u r blind and have no common sense, how the fuck can e be in cd? they are like 2 light years apart, this is literally a visual problem, not some quantum bullshit
BRO U DONT NEED TO BE A GENIUS TO KNOW E IS NOT IN CD
THEY ARE LITERALLY 2 LIGHT YEARS APART
AND IF U MEAN BD, THEN U DONT NEED TO KNOW, BECAUSE IT DONT EVEN MATTER, THE PROCEDURE DOESNT EVEN CHANGE
The way the equations are set up, we would get a negative value for x/y if E was on BC, which we don't.
@@jaspermcjasper3672 Why? That's a basic fact of geometry and not unique to this particular problem.
I solved the problem in a totally different way, using the sin and cos laws to complete each triangle. It definitely felt like a less clever solution but it was super satisfying and a really great way to review a bunch of elementary trig. Definitely keeping this problem in my back pocket for students!!
Set the two equal lengths = 1 and solve for the line intersection point coordinates instead.
Hi,
10:33 : missing "and that's a good place to stop".
I did the same construction of E, and used an angle sum formula, but more simply for both parts.
As no lengths or areas are required, without loss of generality, define AE=1. By well known ratios for 30°-60°-90° and 45°-45°-90° triangles, EC=√3 and ED=1. DC=EC-ED=√3-1, but BD=DC, so BD=√3-1. EB=ED-BD=1-(√3-1)=2-√3. I'm sort of synced up here with Dr Penn at 6:15, same ratios expressed differently.
Defining φ=∠EAB, tanφ=(2-√3)/1.=2-√3. I suspected φ=15°, and a quick check on a calculator indicated this was worth pursuing. Rather than use a half angle formula and have to untangle nested square roots, I opted to work up and calculated tan(2φ)=2tanφ/(1+tan²φ) = 2(2-√3)/(1+(2-√3)²) = (4-2√3)/(1+4+4√3+3) = (4--2√3)/(8-4√3) = (4--2√3)/2(4--2√3) = ½. So 2φ is either 30° or 210°, and only 30° makes sense for the diagram, so indeed φ=15°. Clearly φ+θ=45° from ⊿ADE, so θ=30°.
And IMHO, that's a good place to stop.
Impressed by the pure geometry solutions posted. Like several others, I was drawn to the sine rule approach. Using it on triangle ACD to show that |AD|=|AC|/Sqrt[2] and then applying it to triangles ABD and ABC and eliminating sine of angle ABD leaves sin(theta)=sin(theta+15)/Sqrt[2]. Rather than hammering this out with addition formulae and tan or cot, in this case (particularly recalling sin(45) =1/Sqrt[2]), theta is 30 degrees by inspection.
Much easier by using law of sines. For the larger triangle, Sin (a+15)/2x = sin 30/z and using the smaller triangle, sin 45/z = sin a/x. Simplfying, we get inverse cot of 1.732 which is 30 degrees. ATAGPTS
I had guys like this who put me off maths. The solution is much simpler.
Zwięźle wytłumaczone bez niepotrzebnego rozwlekania.
Can we solve this by drawing a parallel line to BC passing through point A and then using supplementary and complementary angle rules for parallel lines..I am getting stuck though idk
I was thinking like you but not yet successful.
@@rarhs Cleanest shortest geometric solution posted by @papafragen above - Uses 1 circle construction followed by Thales theorem and isoceles theorems...done.
My only marginally different solution was to set the x in the video to 1, without loss of generality, which then immediately gave me tan 60° = sqrt(3) = (y + 2)/(y + 1), and therefor y = (sqrt(3) - 1) / 2.
This then implied that atan(angle EAB) = y / (y + 1) = 2 - sqrt(3), and therefor angle EAB = 15°, and therefor theta = 45° - 15° = 30°.
Just using pitagoras and the half equilateral u can show AB^2 = 2x^2 which implies that AB is the tangent to ADC circumcircle, so BAD = 30.
what i did was to find the angle EAB, which is arctan(y/(x+y)) which comes out to 15°, and then just subtract that from 45°; by that method you don't have to rationalise denominators etc anywhere
I used sine laws
theta = 30
Michael, you are the greatest master of using colored chalk. Nobody seemed to notice that you incorrectly applied a parenthesis at 9:15.
Actually urnoob5528 noticed about 16 hours before you. But the primitive layout of UA-cam comments makes it pretty difficult to notice such things.
From the law of sines, x being the unknown, sin(x+15)=sin(x)/sin(45)
=> Dividing both sides by cos(x), the equation becomes tan(x)/sin(45)=cos(15).tan(x)+sin(15)
=> tan(x)=sin(15).sin(45)/(1-cos(15)sin(45))
Finally, tan(x)=1/sqrt(3) and x=30
This problem can be solved without any construction. As we see that AD is a median of triangle ABC it divides the triangle into two with equal areas. Since angle CAD is 15, using sine formulas for area of the triangles in two different ways and remembering that angle ABC is (180 - (theta + 45)) we get four equations for the area of triangles. Which can be solved very easily for tan theta. Without so much of mathematical jugglery.
Gentile e perspicuo Michael, se posso permettermi, ti dico: assegnato un valore unitario a BD=DC, applicando il teorema dei seni prima al triangolo ACD per ricavare AD e poi il teorema di Al-Kashi (Carnot) al triangolo ADB, per ricavare AB, applicato nuovamente il teorema dei seni al triangolo ABD, l'angolo theta si ricava molto più facilmente della tua pur pregevole prolusione. Un caro saluto.
I just started on the problem and then asked AI what the answer is. She told me in about 3 seconds so that is how I am going to work the problem. I will never have to do this problem in my everyday life and I really need to mow the yard and do some other things so I really like asking AI. After all, isn’t that why we even have computers in the first place to make our lives easier. I need to figure some income tax info too so I think I will use TurboTax and then I will have time to also do the grass edging after I mow.
Drop a perpendicular BH to AC and observe that triangle BHD is equilateral. Notice that angle ADH is 15 degrees, so that triange ADH is isosceles and AH = DH=BD. Hence, points A, B, D lie on the circle centered at H. The inscribed angle BAD then is equal to one half of the 60 degree central angle BHD, that is, 30 degrees.
DAC = 180° - DBA = 180° - 45° = 135°
ACD = 180° - DAC - CDA = 180° - 135° - 30° = 15°
ACB = 180° - CBA - BAC = 180° - 30° - BAC = 150° - BAC
ACB = ADB + ACD = ADB + 15°
BAD = 180° - DBA - ADB = 180° - 45° - ADB = 135° - ADB
Three Equations with using BAD = BAC:
(1) ACB + BAC = 150°
(2) ACB - ADB = 15°
(3) BAC + ADB = 135°
(1) - (2) BAC + ADB = 135°
I just realized that D should be exactly in the middle of AB when I let the video play for 5 seconds :D I was just wondering why my system of equations is underdetermined.
The pure geometry method is much more satisfying... and if you are lucky enough to construct the perpendicular from B to AC (call the new point E) and then join ED... it all falls nicely into place. Mind you, I did flail around for a while before trying that.
After reading some of the comments, I’m thinking pure luck I got the 30 degree answer. Geez I’m glad I will never have to be in a geometry class or a calculus class ever again. And my grandkids can go online for help.
At 9:17 you wrote (3+3^1/2*tanø) instead of writing (3+3^1/2)*tanø, which is correct
Using sine rule is perhaps more easy since AD is common for 2 triangle and BD is CD . So we get equation in sin(Theta)
Since bdc is a straight line,angle abd is equal to 180-(45+30)=105 degrees
So angle bad is equal to 180-(105+45)=30 degrees
It is a very simple problem.
Aight... Math time. The triangle ADC can easily give the three angles as 135 + 30 + 15, and the triangle ABD has angles theta, 45, and 180-45-theta.
Now let's use the law of sines. For any triangle with corners ABC, sin(A)/BC = sin(B) / AC = sin(C) / AB
Since AD is shared between the two triangles, this simplifies to these two equations. Let y = AD and x = BD = DC
sin(theta) / x = sin(180 - 45 - theta) / y
sin(15) / x = sin(30) / y
since sin(180-theta) = sin(theta):
x = y*sin(theta) / sin(45 - theta)
x = y*sin(15) / sin(30)
Combine both, and...
y*sin(theta) / sin(45 - theta) = y*sin(15) / sin(30)
sin(theta) / sin(45 - theta) = sin(15) / sin(30)
theta = 15
A Simple way is to draw a height line BE, the height is the same as BD and DE (because BEC on the same circle, C is the center). So angle EAD equals angle EDA. So DE=EA=EB. So angle CAB=45 so angle dab=30
My math teacher in high-school asked us how does on solve math problems? But nobody could answer including the teacher. Many years later, helping old people to use their phones and young how to do math, I realised the solution. The secret using phones are the same as solving math problems. It is very simple: Do something regardles what. ( don't just sit there and look at it ) . It sounds silly but when you do something it may be wrong or not help but that only means do something else regardless what. This way you can solve all problems. I am now 82 year old retired technical director, I tell you this is true.
How do we know (without drawing it out) that E lies outside BC?
Bit easier using Sine Rule. 10 lines max.
22 degree rake
Using law of sines:
tan ? = (sin45*sin15)/(1-sin45*cos15)
tan ? = 1/sqrt(3)
I used a geometric approach using basic gymnasium level geometry
set AE=1;DE=1;AC=2;CE=3^-2;BE=2-3^-2;EAB=15;BAD=30
that bottom left angle is 90+15 degrees = 105 degrees. Total degrees are 180, so (using the 45 degree and 30 degree ratio) the top angle must be 45 degrees total and the unknown angle must be 2/3 of the 45 degrees or 30 degrees. It's been a while since I delved into geometry so I've forgotten much of the nomenclature, but not the process.
Trigonometry is for losers....(kidding).
3x2(9yz)4a
Let b = angle EAB, d = angle EAB, C = angle EAC, so theta = d - b.
tan(c) = tan(60) = sqrt(3).
tan(d) = tan(45) = 1.
tan(c) - tan(d) = tan(d) - tan(b), so tan(b) = 2*tan(d) - tan(c) = 2 - sqrt(3).
tan(theta) = tan(d - b) = (tan(d) - tan(b)) / (1 + tan(b)tan(d)) = ... = 1/sqrt(3) = tan(30).
I don't get how you neglect the option of y=0 base only on the drawing
Determine angle measures in ADC
Use law of sines
Convert to tangent
theta = 30
is a good place to stop?
I mean D is the center of circle
...and that's a good place to stop!
At 9:18 the closing paranthesis goes to the wrong place.
This video got me to thinking about other contructable whole numbered angles. I tried google but couldn't find any site that has a complete list of the possible angles that can be constructed. (1°~180°)
not enough info in the thumbnail
Faster way to solve:
B=135-θ (angle sum of triangle)
∠DAC=15 (angle sum of triangle)
Set BD=CD=1 unit
AD=0.5csc15 (sine law)
2sin(135-θ)sin15=sinθ (sine law)
tanθ=(√(2)sin15)/(1-√(2)sin15)
tanθ=1/√3
θ=30
Let T = theta. AD is common side of 2 triangles and BD = DC, so *sine rule* can solve this problem in the simplest way. Angle B = 180 - 45 - T = 135 - T and Angle DAC = 45 - 30 = 15.
Therefore, using *sine rule* sin(135-T) / sin T = sin(30) / sin(15) = (1/2) / [(√3-1) / 2√2] = (√3+1)/√2
sin(135-T) = sin(45+T) = (sin T + cos T)/√2 = sin T (1 + cot T)/√2
Therefore, (√3+1)/√2 = sin(135-T) / sin T = (1 + cot T)/√2, so cot T = √3, so *theta = T = 30 deg*
Ok. I’m no math genius. And by the time you were done, I was lost and confused. But I realized that the angle was thirty degrees. And I did that using the calculator on my iPhone. And that’s what you ended up with, while making me feel like how I felt in math classes through primary school to college. Blank stare on my face and huh? I mean it took me less than a minute.
Forgot the first step: convert to radians.
From law of sines:
x/sin(45) =1/sin(θ)
x/sin(30) =2/sin(θ+15)
sin(45)/sin(θ) =1/sin(θ+15)
sin(θ)=sin(45) sin(θ+15)
Let a=θ-30
sin(a+30)=sin(45) sin(45+a)
sin(a) cos(30)+cos(a) sin(30)=sin(45) (sin(a) cos(45)+cos(a) sin(45))
sin(a) cos(30)+cos(a) sin(30) = 1/2 (sin(a) + cos(a))
2 sin(a) cos(30)=sin(a)
sin(a)=0
-> a=0 -> θ=30
I found a shorter solution. If x = BD = DC, and y = AB, By the Law of Sines in triangles ABD and ABC, sin θ/x = sin 45°/y and sin (θ+15°)/(2x) = sin 30°/y. Dividing both members, we can eliminate easily x and y, and then sin (θ+15°) sin 45° = sin θ. This equation has the obvious solution 30° because sin 30° = 1/2 = (1/√2)² = sin² 45°. You can solve the last equation doing some tricks I could post.
I never recall the exact values of sin 15° and cos 15°, so I did the following.
Recall that sin (θ+15°) sin 45° = sin θ, thus -sin (θ+15°) sin 45° = -2sin θ. Then, cos (θ+60°)-cos(θ-30°) = - 2 sin θ. Expanding the left side, etc. we find 1/2 (3-√3) sin θ = 1/2 (√3-1) cos θ.
Simplifying, tan θ = 1/√3, so θ = atan (1/√3) = 30°.
Isn’t it far easier to just put equations of both lines we know slopes and their intercept. See where they intersect
A "pen to paper" solution would have been far easier to follow.
Agatha Christie moment, B-D and C-D are the same length.
New character who arrives on the next to last page "who did it".
Crap, actually.
No good place to stop?
Just for you: it's a good place to stop.
amazing problem
tan(ô)=(sin15)(sin45)/((sin30)-(sin15)(sin45)) so ô=30°
Up-to 3-17 may take ur solution.
" Find angle marked x" and not find "missing" angle. It's not missing 🙂☹️
180 - 45 -30 = 105
105 - 90 = 15
45 -15 = 30
30
Way too complicated! This was a 20 second problem. We can apply numerical values to the sides since we have 30˚. AC = 4, AE = 2, BD = DC = √3 -x. ED = x+√3=AE = 2, EB=2x, EAB = arc tan 2-√3 = 15˚, since EAB = DAC = 15˚, therefore BAD = 30˚.
I hate it when the question in the thumbnail isn't the question answered in the video.
it's enough to easily get it from 180-45-30=105 (missing angle)
@@TimothyReeves That is the sum of the two angles at the top, not the missing angle.
And that's a good place to stop.
Not today.
Maybe I just got lucky but the bottom straight line =180deg subtract 45 and 30 deg =105 so ABD is 105 add the 45 deg to that then subtract that from 180 and I get 30 deg
I did my way in like 20 seconds
You gave half info in thumbnail and did not say BD=DC.
Not fair ! 😔
Uncle ji
Total angle
180 - addition of both the angles
Theta +beta
Means
180-75 = 105
105-75= 30
😂😂
Oooh, jalannya ngeliat monitor di bawah toh ternyata.. Hmm...
Sancs
Very very very …long solve
its 15 deg
Impressive
And that's a great place to stop!!! ❤
Clickbait! It's made to look impossible because D being the midpoint is not a given until suddenly it is.
By inspection, angle CDA =135 degrees, so angle DAC =15 degrees. Applying the law of sines yields the dimensions AD & AC with respect to X. Then apply the law of cosine to triangle ABC (with the obtained value of AC, BC = 2X and the angle between them = 30 degrees) to find AB (alternatively: apply the law of cosines to ABD with the obtained value of AD, BD=X and angle BDA=45 degrees). Now apply the law of sines: X/ sin(theta) = AB/ sin(45).
You build the answer about wrong assuming bd=dc
BD=DC
45°=
75.
45
?? Degrees 😂