Question

SOMEONE PLEASE HELP!!!

Answer 1

`z-x=160^\circ-x-z\implies2z=160^\circ\implies z=80^\circ`

By the law of sines, in triangle ABD we have

`\frac{\sin x}b=\frac{\sin(160^\circ-x)}a\implies\frac ab=(\sin(160^\circ-x))/(\sin x)`

and in triangle ABC,

`\frac{\sin80^\circ}a=\frac{\sin20^\circ}b\implies\frac ab=(\sin80^\circ)/(\sin20^\circ)`

By substitution we end up with

`(\sin(160^\circ-x))/(\sin x)=(\sin80^\circ)/(\sin20^\circ)`

`(\sin160^\circ\cos x-\cos160^\circ\sin x)/(\sin x)=(\sin80^\circ)/(\sin^\circ)`

`\sin160^\circ\cot x-\cos160^\circ=(\sin80^\circ)/(\sin20^\circ)`

`\cot x=(\sin80^\circ+\sin20^\circ\cos160^\circ)/(\sin20^\circ\sin160^\circ)`

`x=\cot^(-1)\left((\sin80^\circ+\sin20^\circ\cos160^\circ)/(\sin20^\circ\sin160^\circ)\right)=10^\circ`

There's probably some sequence of identities you can apply to the right side to somehow end up with
`\cot x=\cot10^\circ` but I'm not seeing it right away... I just used a calculator at this point.