Question

Four identical mirrors are made to stand vertically to form a square arrangement as shown in a top view. A ray starts from the midpoint M of mirror AD and after two reflections reaches corner D. Then, angle $\theta$ must be

• A. $\tan {}^{-1}\left(0.75\right)$
• B. $\cot {}^{-1}\left(0.75\right)$
• C. $\sin {}^{-1}\left(0.75\right)$
• D. $\cos {}^{-1}\left(0.75\right)$

Answer 1

The correct option is B $\cot {}^{-1}\left(0.75\right)$

$\alpha =90°-\theta \therefore \beta =90°-(90°-\theta )=\theta \gamma =90°-\beta =90°-\theta \delta =90°-\gamma =90°-(90°-\gamma )=\gamma \tan \theta =\frac{AP}{AM}$

$\therefore BP=a-AM\tan \theta =a-\frac{a}{2}\tan \theta$ $[a⟶$side of the square$]$

$\therefore \tan \theta =\frac{BP}{BQ}$ $($from $\Delta BPQ)$

$\therefore$In $\Delta QCD,$

$\tan \delta =\tan \theta =\frac{a}{QC}=\frac{a}{a-BQ}=\frac{a}{a-\frac{(a-\frac{a}{2}\tan \theta )}{\tan \theta }}⟹\tan \theta =\frac{a\tan \theta }{3\frac{a}{2}\tan \theta -a}⟹\tan \theta =\frac{\tan \theta }{\frac{3}{2}\tan \theta -1}⟹\frac{3}{2}\tan \theta -1=1⟹\frac{3}{2}\tan \theta =2⟹\tan \theta =\frac{4}{3}\therefore \theta ={\cot }^{-1}\left(0.75\right)$