Question

AB is a vertical pole. The end A is on the level ground. C is middle point of AB and P is a point on the level ground. The portion CB subtends an angle $\beta$ at P. If AP = n AB, then tan $\beta$ =

• A. $\frac{n}{n^2-1}$
• B. $\frac{n}{n^2÷1}$
• C. $\frac{n}{2n^2+1}$
• D. none

Answer 1

Answer: (C)

$\frac{n}{2n^2+1}$

Given:

AB is vertical pole with A at ground. C is midpoint of AB

P is another point in ground and Distance of AP= nAB

Let $AB=X$

So, $AC=\frac{X}{2}$

$AP=nX$

Given that Beta is the angle subtended by BC at P

Assume Alpha is the angle subtended by AC, such that AB subtends an angle of (Alpha +Beta) at P

Therefore, tan $\left(\alpha \right)=\frac{\left(X/2\right)}{nX}=\frac{1}{2n}$

$tan(\alpha +\beta )=\frac{X}{nX}=\frac{1}{n}$

We know the formula: $tan(x+y)$ = $\frac{[tanx+tany]}{[1-tanx\ast tany]}$

Therefore, $\frac{1}{n}$= $\frac{[tan\beta +1/2n]}{[1-(tan\beta \ast 1/2n\left)\right]}$

After solving the above equation,

$tan\beta =\frac{n}{(2n^2+1)}$