# Find angle ADB in this isoceles triangle given some extra information

• chwala
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
Trigonometry
Question;

My take,

I have,

then
using sine rule;

##\dfrac{x}{x+y} = \dfrac{\sin 20^{\circ}}{\sin 80^{\circ}}##

##\dfrac{x}{x+y} =0.347##

##x=3.47## then ##y=6.53##.

then,

##BD^2=3.47^2+10^2-(2×3.47×10×\cos 20^{\circ})##

##BD= 6.842##

...

##10^2=3.47^2+6.842^2-(2×3.47×6.842 ×\cos m)##

##41.15=-47.4696\cos m##

##\cos m = -0.866##

##m= 150^{\circ}##

there could be a better approach...just came across this question today.

dipcario34
## \frac {x}{x+y} = \frac {\sin 20^\circ}{\sin 80^\circ} = \frac {\sin (160^\circ-m)}{\sin m} ##

chwala
Gavran said:
## \frac {x}{x+y} = \frac {\sin 20^\circ}{\sin 80^\circ} = \frac {\sin (160^\circ-m)}{\sin m} ##
@Gavran following your approach, i am ending up with ##\tan m = -0.5773##

I'll post steps later once I get my hands on laptop...(a bit tedious as you have to use sine expansion) as I am using phone to type.

We then end up with,

##m= -30^0##

...then repititive every ##180^0## for tan, implying

##m= -30^0+180^0=150^0##

Unless am missing something...that's how I look at your approach...

Last edited:
Gavran
Gavran said:
##\displaystyle \frac {x}{x+y} = \frac {\sin 20^\circ}{\sin 80^\circ} = \frac {\sin (160^\circ-m)}{\sin m} ##
I don't know if this will help OP, but ##\displaystyle\ \frac {\sin 20^\circ}{\sin 80^\circ}\ ## can be simplified to ##\displaystyle\ {2\sin 10^\circ}\ .## Use the double angle identity for ##\sin 20^\circ## and that ##\displaystyle \sin 80^\circ = \cos 10^\circ## .

Or simpler: Use a construction to find ##\displaystyle\ \sin \frac{20^\circ}{2} \ ## directly.

Gavran, Lnewqban and chwala
SammyS said:
I don't know if this will help OP, but ##\displaystyle\ \frac {\sin 20^\circ}{\sin 80^\circ}\ ## can be simplified to ##\displaystyle\ {2\sin 10^\circ}\ .## Use the double angle identity for ##\sin 20^\circ## and that ##\displaystyle \sin 80^\circ = \cos 10^\circ## .

Or simpler: Use a construction to find ##\displaystyle\ \sin \frac{20^\circ}{2} \ ## directly.
Smart move man @SammyS

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