The principal section of a glass prism is an isosceles triangle $A B C$ with $A B=A C$. The face $A C$ is…
- $30^{\circ}$
- $36^{\circ}$
- $60^{\circ}$
- $72^{\circ}$
Solution

$ \begin{array}{llrl} & i_1 =90^{\circ}-\left(90^{\circ}-A\right)=A \\ \text { and } \alpha & =90^{\circ}-21_1=90^{\circ}-2 A \\ \therefore & i_2 =90^{\circ}-\alpha=90^{\circ}-\left(90^{\circ}-2 A\right)=2 A \\ \therefore & \beta =90^{\circ}-i_2=90^{\circ}-2 A \end{array} $ From the geometry of the figure $ \begin{aligned} & A+2 A+2 A=180^{\circ} \\ & \therefore \quad A=36^{\circ} \end{aligned} $
Asked in: AP EAMCET 2004