Question

In the figure 2.56, $\angle$X = 62$°$, $\angle$XYZ = 56$°$. If YI and ZI are the bisectors of $\angle$XYZ and $\angle$XZY respectively of $∆$XYZ, find $\angle$IZY and $\angle$YIZ.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}\angle \mathrm{X}=62°\mathrm{and}\angle \mathrm{XYZ}=56° \\ \mathrm{Now},\angle \mathrm{XYI}=\angle \mathrm{IYZ}=\frac{\angle \mathrm{XYZ}}{2}=\frac{56°}{2}=28°(\because \mathrm{YI}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector})...\left(\mathrm{i}\right) \\ \mathrm{and}\angle \mathrm{XZI}=\angle \mathrm{IZY}=\frac{\angle \mathrm{XZY}}{2}(\because \mathrm{ZI}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}) \\ \mathrm{In}△\mathrm{XYZ},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{X}+\angle \mathrm{Y}+\angle \mathrm{Z}=180° \\ \Rightarrow 62°+56°+\angle \mathrm{Z}=180° \\ \Rightarrow \angle \mathrm{Z}=62° \\ \therefore \angle \mathrm{IZY}=\frac{\angle \mathrm{XZY}}{2}=\frac{62}{2}=31°....\left(\mathrm{ii}\right) \\ \mathrm{In}△\mathrm{IYZ},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{IYZ}+\angle \mathrm{IZY}+\angle \mathrm{YIZ}=180° \\ \Rightarrow 28°+31°+\angle \mathrm{YIZ}=180°\left[\mathrm{from}\right(\mathrm{i})\mathrm{and}(\mathrm{ii}\left)\right] \\ \Rightarrow \angle \mathrm{YIZ}=121° \end{gathered}$$