Mathematics - Heights and Distances Question with Solution | TestHub

We have, $16^{{\sin }^2\theta }+16^{{\cos }^2\theta }=10$ $\Rightarrow 16^{{\sin }^2\theta }+16^{1-{\sin }^2\theta }=10$ $\Rightarrow x+\frac{16}{x}=10$, where $x=16^{{\sin }^2\theta }$ $\Rightarrow x^2-10x+16=0\Rightarrow x=2,8$ $\therefore 16^{{\sin }^2\theta }=2,8\Rightarrow 2^{4{\sin }^2\theta }=2,2^3$ $\Rightarrow 4{\sin }^2\theta =1,3\Rightarrow {\sin }^2\theta =\frac{1}{4},{\left(\frac{\sqrt{3}}{2}\right)}^2$ $\Rightarrow \sin \theta =\frac{1}{2},\frac{\sqrt{3}}{2}\Rightarrow \theta =\frac{\pi }{6},\frac{\pi }{3}$
Let the altitude of the sun be $\theta$. Then, $$\begin{array}{rlrl}& & \tan \theta & =\frac{h}{h/\sqrt{3}}\Rightarrow \tan \theta =\sqrt{3}\\ \Rightarrow & & \theta & =\pi /3\Rightarrow \theta =2\alpha \end{array}$$