Question

If $\sec \theta \cdot \sin ({36}^o+\theta )=1$, find the value of $\theta$ so that $\theta$ and ${36}^o+\theta$ are acute angles.

Answer 1

$\begin{array}{c}wehave\\ \Rightarrow \sec \theta .\sin \left({36}^{\circ }+\theta \right)=1\\ \Rightarrow \sin \left({36}^{\circ }+\theta \right)=\cos \theta \\ \Rightarrow \sin \left({36}^{\circ }+\theta \right)=\sin \left({90}^{\circ }-\theta \right)\\ \Rightarrow \left({36}^{\circ }+\theta \right)=\left({90}^{\circ }+\theta \right)\\ \Rightarrow 2\theta ={54}^{\circ }\\ \therefore \theta ={27}^{\circ }\\ Now,\\ {36}^{\circ }+\theta ={36}^{\circ }+{27}^{\circ }\\ ={63}^{\circ }\\ Hence,thevalueof\theta ={27}^{\circ }and{36}^{\circ }+\theta isnotanacuteangle.\end{array}$