The 80-80-20 Triangle Problem, Solution #3
Let ABC be an isosceles triangle
|Contact| |Front page| |Contents| |Geometry| |Up|
Copyright © 1996-2018 Alexander Bogomolny
This is a trigonometric solution (Solution 2 from [Knop]).
Denote the unknown angle CED as x. Then
Apply the Law of Sines in ΔCDE:
CE : CD = sin (160° - x) : sin x.
And in ΔBCE,
CE : BC = sin 80° : sin 40° = 2cos 40°.
For, sin 2α = 2sin α · cos α.
Since ΔBCD is isosceles, CD = BC, which allows us to write a trigonometric equation:
sin (160° - x) : sin x = sin 80° : sin 40°.
This we are going to solve presently.
Using
sin (20° + x) = 2cos 40° · sin x = 2cos (60° - 20°) · sin x.
Using the addition formula for sine and the one for cosine,
sin 20° · cos x + cos 20° · sin x = cos 20° · sin x + √3sin 20° · sin x,
which simplifies to
sin 20° · cos x = √3sin 20° · sin x,
or,
ctg x = √3.
From which x = 30°.
Reference
- C. Knop, Nine Solutions to One Problem, Kvant, 1993, no 6.
The 80-80-20 Triangle Problem
- Solution #1
- Solution #2
- Solution #3
- Solution #4
- Solution #5
- Solution #6
- Solution #7
- Solution #8
- Solution #9
- Solution #10
- Solution #11
- Solution #12
|Contact| |Front page| |Contents| |Geometry| |Up|
Copyright © 1996-2018 Alexander Bogomolny
71544510