Question

At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $\frac{5}{12}$. On walking $160$ metres towards the tower, the tangent of the angle of elevation is $\frac{3}{4}$. The height of the tower is equal to

• A. $150$m
• B. $180$m
• C. $200$m
• D. None

Answer 1

The correct option is A $150$m

Let $h$ be the height of the tower.

Given, at 1st point on the ground, $\tan \alpha =\frac{5}{12}$ ...given
$\therefore \tan \alpha =\frac{h}{x+160}$

Now we can write $\frac{5}{12}=\frac{h}{x+160}$

$12h=5x+800$ ...(1)
At second point on the ground, $\tan \beta =\frac{3}{4}$ ...given
$\tan \beta =\frac{h}{x}$

where, $x$ is the distance of 2nd point on the ground from tower
Now we can write $\frac{h}{x}=\frac{3}{4}$
$\therefore x=\frac{4h}{3}$ ...(2)
Put the value of $x$ from eq 2 in eq 1, we get $h=150m$