Question

From the top of a tower, of $100 \, m$ height, the angles of depression of two objects, $\frac{200}{\sqrt{3}}m$ apart on the horizontal plane on a line passing through the foot of the tower and on the same side of the tower are $45^{o}-A$ and $45^{o}+A.$ Then angle $A$ is equal to

• A. $15^{o}$
• B. $35^{o}$
• C. $22\frac{1^{o}}{2}$
• D. $45^{o}$

Answer 1

Answer: (A)

Let $OP$ be tower and objects are placed at $A \, \& \, B$ , then
The distance between the objects
$=100\left[c o t \left(\left(45\right)^{o} - A\right) - c o t \left(\left(45\right)^{o} + A\right)\right]$ $\left\{\right.from\Delta OPA \, \& \, \Delta OPB\left.\right\}$
$=100\left(\frac{1 + t a n A}{1 - t a n A} - \frac{1 - t a n A}{1 + t a n A}\right)$
$=100\left(\frac{\left(1 + tan A\right)^{2} - \left(1 - tan ⁡ A\right)^{2}}{1 - \left(t a n\right)^{2} A}\right)$
$=100.\frac{4 tan A}{1 - t a n^{2} A}=200tan ⁡ 2 A$
$\therefore \frac{200}{\sqrt{3}}=200tan 2 A$
$\Rightarrow tan 2A=\frac{1}{\sqrt{3}}$
$\therefore 2A=30^{o}\Rightarrow A=15^{o}$