<s>[INST] Solve for x $ \sin 30° \sin x \sin 10° = \sin 20° \sin ({80°-x}) \sin 40° $ $ \sin 30° \sin x \sin 10° = \sin 20° \sin ({80°-x}) \sin 40° $
I tried transformation formulas , $ 2\sin a \sin b $ one. I know the value of sin 30° but what about others?
Original problem
In triangle ABC, P is an interior point such that $ \angle PAB = 10°. \angle PBA = 20° PAC = 40°  \angle PCA = 30° $ then what kind of triangle it is ?  I solved it till I got stuck here.
 [/INST] Assume AB = 1. 

Apply sine law to PAB to get $PB = 2 \sin 10^0$.
Apply sine law to PAC to get $PB = 2 \sin 20^0$ and $PC = 4 \sin 20^0 \sin 40^0$.
Apply cosine law to PBC to get $BC$ in terms of those angles. Note that $\cos 100^0 = - \sin 10^0$.
$BC^2 = 4 \sin^2 10 + 16 \sin^220^0\sin^2 40^0 + 16 \sin 20^0 \sin 40^0 \sin^210^0$
Suggestion:- Convert all angles in $BC^2$to $\sin 10^0$ (or $\sin 20^0$) by compound angle formula. Hope that the result is 1. 
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