A vertical tower standing at $O$ has marks $P, Q, R, S$ at heights of $1 m, 2 m, 3 m$ and $4 m$ from the foot $O$ and $A$ is a point on the horizontal plane through $O .$ If $P Q$ and RS subtend angles $α$ and $β$ respectively at A where $O A = 2 m$ then $\cos (α + β) =$

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$5 / \sqrt{26}$

$24 / \sqrt{650}$

$23 / \sqrt{650}$

$1 / \sqrt{26}$

answer is C.

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Detailed Solution

$\begin{aligned} (A P)^{2} = 4 + 1 = 5 \\ (A Q)^{2} = 4 + 4 = 8 \\ (A R)^{2} = 4 + 9 = 13 \\ (A S)^{2} = 4 + 4 = 16 \end{aligned}$

$\begin{array}{l} \cos α = \frac{8 + 5 - 1}{2 \sqrt{8 \times 5}} = \frac{3}{\sqrt{10}} \Rightarrow \sin α = \frac{1}{\sqrt{10}} \\ \cos β = \frac{20 + 13 - 1}{2 \sqrt{20 \times 13}} = \frac{8}{\sqrt{65}} \Rightarrow \sin β = \frac{1}{\sqrt{65}} \\ \cos (α + β) = \frac{3 \times 8}{\sqrt{650}} - \frac{1 \times 1}{\sqrt{650}} = \frac{23}{\sqrt{650}} \end{array}$