Find angle ADB in this isoceles triangle given some extra information

[chwala]

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.

 

[Gavran]

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

 

[chwala]

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

@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...

 

[SammyS]

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

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.

 

[chwala]

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