Mathematics - Properties of Triangles Question with Solution | TestHub

We have, $\tan \frac{X}{2}+\tan \frac{Z}{2}=\frac{2y}{x+y+z} \& s=x+y+z=$perimeter of a triangle  

Apply half-angle formula and $x+y+z=s$ then, 

$\Rightarrow \frac{∆}{s\left(s-x\right)}+\frac{∆}{s\left(s-z\right)}=\frac{y}{s}\Rightarrow ∆=\left(s-x\right)\left(s-z\right)$

$\Rightarrow {∆}^2=s\left(s-x\right)\left(s-y\right)\left(s-z\right)={\left(s-x\right)}^2{\left(s-z\right)}^2$

$\Rightarrow y^2=x^2+z^2\Rightarrow \angle Y=90°$

$\Rightarrow \angle Y=\angle X+\angle Z$

An option $\left(B\right)$ is correct.

Now, $\frac{x}{y+z}=\frac{\sin {\displaystyle X}}{\sin {\displaystyle Y}{\displaystyle +}{\displaystyle \sin }{\displaystyle Z}}=\frac{{\displaystyle \sin X}}{{\displaystyle 1+\sin \left(\frac{\pi }{2}-X\right)}}=\frac{2\sin {\displaystyle \frac{X}{2}}\cos {\displaystyle \frac{X}{2}}}{2{\cos }^2{\displaystyle \frac{X}{2}}}=\tan \frac{X}{2}$

An Option $\left(C\right)$ is correct.