Learn new techniques to manipulate this complex problem | Find angle X | Math Olympiad
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- Опубліковано 15 сер 2023
- Learn how to find the angle X in a triangle. Important Geometry, Trigonometry, and Algebra skills are also explained: Exterior Angles; Law of Sines; Law of Cosines; Trigonometry. Step-by-step tutorial by PreMath.com
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Impressive, informative, and entertaining. Well done sir
Glad you enjoyed it!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍
Thank you!
The geometrical solution that I came up with starts by drawing an isosceles triangle underneath with its base as AB and two sides of length AC, call it ABE. It creates two additional isosceles triangles, ACE and BDE. From that x=40 can be derived.
Sorry. I'd accidentally connected a line which then implied an assumption that happened to be a correct one. So better to attack it from the other direction, ie rather than drawing triangle ABE, make the angle DBE of 20 degrees then extend CD that intersects it in E. That makes an isosceles triangle (base angles=50) CEB. It's apex is 80 thereby making the isosceles triangle BDE. Drawing a circle with center C and radius AC then is used to pin down the values. A few auxiliary lines need to be added. The other way with triangle ABE might result in a solution however I did not see it.
AC must be greater than AD. If AC=AD=DB, then CDB must equal 120 degrees (draw a circle with center at A and radius AC, for ABC=30, ACB=90 when BC is tangent to the circle, and the tangent to the circle that intersects AB at B makes an angle of 30 degrees). Hence in that case CDB is greater than 100. If AC is less than AD, then that circle would be further from B, hence BC would not intersect the circle when CBA=30 and hence not conform to the diagram. Therefore AC must be greater than AD, and therefore making the isosceles triangle possible. However, as I mentioned in my comment to my post, the diagram described in the post does not produce a solution as far as I can tell. The diagram described in my comment above does.
The solution ends up being a bit involved. First start from the diagram I gave in my comment above that draws a 20 degree angle. From E draw the pendicular to AB that continues up to the parallel to AB through C. Call the intersection with the parallel 'G'. Lay out a length along the parallel from G equal to CG making a length CH twice the length of CG. Consequently making two congruent triangles ECG and EHG. Draw BH. Now BE=EH=EC. Therefore C, H, and B lie on a circle with center E. Triangles ECG and EHG are 80-10 right triangles, therefore angle CBH is 10 degrees, making EBH 60 degrees. Angles EBH=EHB (isosceles triangle) therefore triangle EBH is equilateral and BH=BE=AC. Therefore angle ABH=BAC=x=40 degrees.
If there is a geometrical solution, please make another video for this nice problem!
Insightful...! 🙂
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A lovely lesson with a pleasing result.
Glad you enjoyed it
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Always three steps ahead, Professor!👍🥂❤️
Excellent!
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Thanks for video.Good luck sir!!!!!!!!!!
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Very good, and at last sin without calculator! Just by shortening!
Geometrical solution: 1) Flip ∆CDB with axle CB getting point E outside ∆ABC 2) Flip ∆CDE with axle CD getting point F outside ∆ABC 3) Since ∠DCE = ∠DCF is 100°, ∠CDB is 100°, so FC is parallel with AB. DB=EB=ED, ED=FD. Make a parallel line with FD from C to point G on the line AB, so CG = FD, ∠CGA = ∠FDA = 40°and CA=CG, finally, x = 40°
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I can't imagine that. Would you describe the solution bit more detail, please?
Sorry, I got typo, the 3rd step is Making parallel line with FD, so FDGC is Parallelogram@@n_eros
Clever!
In those triangles:
- Side 'a' is common
- Side 'm' is same value
Sine rule: (Triangle at right)
a/sin 30° = m / sin 50°
a/ m = sin 30°/sin 50° = 0,6527
Sine rule (Triangle at left)
a / sin x = m / sin (30+50)
a/m = sin x / sin 80°
Equalling:
sin x / sin 80° = 0,6527
x = 40° ( Solved √ )
Surely it's simpler to draw a line front D parallel to AC and then a line from C to the join this new parallel line? You then have a quadrilateral with two opposite angles summing to 100 degrees and then 2X = 80 and x = 40? Or am I missing something?
I did get a different result and perhaps wrong, but hear me out and find if/where I made a mistake. I was trying a geometrocal approach. I took two of suggested triangles and made a parallelogram with them. That way my two slanted sides (one which is marked here at the angle X and the other one going from 30 deg angle) are the same - as they should be in a parralelogram. The triangle side defined with the 50 and 30 angles would be the horizontal base of the said parallelogram. Then the marked side from the 30 angle would be half of the parallelograms diagonal. This all gave me an isoscales triangle (slanted parallelogram's side is equal to half of the diagonal) and got that X is 20 (2x80 deg at the base and 20 on the top). If X is 40 deg, then in my geometrical representation (parallelogram) I don't see how the lines marked in the triangle can be the same. How come I got here and cant find where/if I am mistaking. Is my result valid?
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At first, I determined the value of sin(x) with a calculator according to the expression also given in the video. After getting x, I proved the solution to be exact in the same way you did it. Do you think there is a change to see immediately how to manipulate the expression to get the exact result? Or is it just the sum of experiences?
Best regards from Germany
Great!
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40 Answer
Will use the law of Sines, Construct a 30-60-90 right triangle, Pythagorean Theorem, and finally
the Law of Cosines.
For triangle BCD, let the length of BC=4. (I chose 4 since later on, I will construct a 30-60-90. Hence the other
sides are 2, and 2sqrt 3)
Using the law of sines BD =3.111
Drop a perpendicular from C to line AD to point P forming a 30,60, 90 right triangle hence CP= 2. PB
hence is 2 sqrt 3.
Draw a line from C to A (similar to the figure above) of length 3.111. We still do not know the angle. But
we know that ACP is a right triangle with sides 3.111 and 2. Using Pythagorean Theorem AP = 2.38292278515.
Hence line AP = 2.38292278515 + 2 sqrt 3 = 5.847
Hence the three sides of Triangle ABC are 3.111, 4, and 5.847
Now there are three sides, and hence the Law of Cosines can be used to find the inner angles
Angle x = 40 degrees
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There is a slight error, it is not true that sin a = sin b only implies that a = b, there is a second solution that has to be considered, in other words sin a = sin b could also imply that 180-a = b, It just so happens that in this case the second solution doesn't hold.
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is there any geometrical solution?
Yes, there is!
I'll try to put it pretty soon.
Thanks for asking. Cheers
@@PreMath please put it as soon as pissible bcz i try, but cant solve.
CD/Sin30 = DB/Sin50; CD/SinX = AC/Sin80. SinX = (CD*Sin80)/AC = DB*Sin30*Sin80/AC*Sin50 = 0.643. X = 40
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Calculator is allowed? kkkkk Very good explanation teacher!
I probably cheated, but I just assigned a value to the two congruent sides and then used the law of sines to find value x. I let AC and BD = 10. Using law of sines with my numbers, CD = 6.527036. Angle ADC = 80 degrees. AC = 10. Using law of sines for angle x, angle x = 40 degrees.
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a:b:c=sin30:sin 50:sin 100=sin30:sin50:sin80, a:b=sin x:sin80,a/b=sin30/sin50=sin x/sin 80, x=arcsin(sin 30 xsin 80÷sin 50)=40.😅
Excellent!
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Easy one. Can you show more difficult one.
Let's focus on triangle CDB only. Extend CD downwards and make a segment DE so that angle EBD is 20 deg. Now triangle CEB is isosceles with angles 50,80,50 and triangle DBE is isosceles with angles 80,20,80. This means DB = CE = EB. Now extent DB to the left and draw segment EA = EB. Now you have isosceles triangle AEB with angles 20,140,20. Now AE is EC and angle AEC = 140 - 80 = 60. So triangle AEC is equilateral. Now if we draw segment AC it will be equal to DB and we actually constructed the original shape. From there we can conclude angle CAD = 40.
Concerning your step EA=EB, it is not known a priori that A lies on the circle with center E and radius EB. Therefore you cannot conclude at that point that EA=EB.
If you start from triangle ABC you only know one angle so you get stuck. In opposite, triangle BCD is fully defined. So from this triangle I work backwards and find point A which is aligned with DB and its distance from point C is same as DB. So I have indirectly constructed ABC starting from BCD.@@jeffreygreen7860
@@georgexomeritakis2793, That is fine. The circle with center at C with radius DB cuts the line DB at A, that is true in the original diagram.. The point is what is the justification for saying EA=EB?
@@jeffreygreen7860 i constructed triangle AEB as isosceles
@@georgexomeritakis2793 , those are two different A's. Creating the isosceles triangle EA'B where EA'=EB does not guarantee A'=A.
Per il teorema dei seni, e lavorando un po', sinx=sin30sin80/sin50=sin302sin40cos40/cos40=sin302sin40=sin40... x=40
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You didn't take into account periodicity. An example is how sin(40°)=sin(150°).
You can still justify your last step. Since angle ADC is 80°, we know that x is less than 100° (angles in triangle). By having this limit on the range, 40° becomes the only angle x satisfying sin(x)=sin(40°), hence you can take your last step.
X=40
40
Great!
x=20
I was right there with you until the very last step. It is not true that sin a = sin b implies a = b. For a complete solution, you must explicitly eliminate the possibility that x = 140 degrees.
It is known from the figure that x is acute so x < 90°.