Question

From the top of a tower, of $100m$ 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
$=100\left[cot\left({\left(45\right)}^o-A\right)-cot\left({\left(45\right)}^o+A\right)\right]$ $\{from\Delta OPA\&\Delta OPB\}$
$=100\left(\frac{1+tanA}{1-tanA}-\frac{1-tanA}{1+tanA}\right)$
$=100\left(\frac{{\left(1+tanA\right)}^2-{\left(1-tanA\right)}^2}{1-{\left(tan\right)}^{2}A}\right)$
$=100.\frac{4tanA}{1-ta{n}^{2}A}=200tan2A$
$\therefore \frac{200}{\sqrt{3}}=200tan2A$
$\Rightarrow tan2A=\frac{1}{\sqrt{3}}$
$\therefore 2A=30^o\Rightarrow A=15^o$