<s>[INST] How to find this angle? $P$ is a point in $\triangle ABC$. If $\angle PAB = 10^\circ$, $\angle PBA = 20^\circ$, $\angle PAC = 40^\circ$ and $\angle PCA = 30^\circ$, find $\angle B$.
 [/INST] Let $\angle PBC=x$. so for the condition $\angle PCB=80^\circ-x$.


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*Apply "trigonometric expression of ceva's theorem"-$ \frac{\sin 10^\circ}{\sin 40^\circ}\times  \frac{\sin x}{\sin 20^\circ}\times \frac{\sin 30^\circ}{\sin(80^\circ-x)} =1$

*From that we have $\frac{\sin x}{\sin(80^\circ-x)}=4\cos 10^\circ \sin 40^\circ=4\cos 10^\circ\cos 50^\circ=2(\cos 60^\circ+\cos 40^\circ)=1+2\cos 40^\circ$

*Now see .........$\sin 60^\circ-\sin 20^\circ=2\cos 40^\circ\sin 20^\circ$. from this equation we have  $1+2\cos 40^\circ=\frac{\sin 60^\circ}{\sin 20^\circ}=\frac{\sin x}{\sin(80^\circ-x)}$. so comparing we can say $x=60^\circ$

*so $\angle B=20^\circ+60^\circ=80^\circ$.
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