Question

The vertical angle A of a triangle is divided into two parts, $\alpha$ and $\beta$ such that $tan\alpha =3tan\beta$ and $\alpha -\beta ={30}^o$

Answer 1

$\tan \alpha =3\tan \beta \Rightarrow \frac{\sin \alpha }{\cos \alpha }=3\frac{\sin \beta }{\cos \beta }$ ...(1)
A) From (1)
$\sin \alpha \cos \beta =3\sin \beta \cos \alpha \Rightarrow -3\sin \alpha \cos \beta +\sin \alpha \cos \beta =3\sin \beta \cos \alpha -3\sin \alpha \cos \beta \Rightarrow -2\sin \alpha \cos \beta =-3\sin (\alpha -\beta )\Rightarrow \sin \alpha \cos \beta =\frac{3}{4}\Rightarrow 4\sin \alpha \cos \beta =3$
B) From (1)
$\sin \alpha \cos \beta =3\sin \beta \cos \alpha \Rightarrow \sin \alpha \cos \beta -\sin \beta \cos \alpha =2\sin \beta \cos \alpha \Rightarrow \sin (\alpha -\beta )=2\sin \beta \cos \alpha \Rightarrow \sin \beta \cos \alpha =\frac{1}{4}\Rightarrow 4\sin \beta \cos \alpha =1$
$\sin \alpha \cos \beta +\sin \beta \cos \alpha =\frac{3}{4}+\frac{1}{4}=1\Rightarrow \sin (\alpha +\beta )=1\Rightarrow \alpha +\beta =90$
As $\alpha -\beta =30$, gives $\alpha =60$
C) $2\sin \alpha =2\frac{\sqrt{3}}{2}=\sqrt{3}$
D) $4\tan \frac{\alpha }{2}=\frac{4}{\sqrt{3}}$