Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them. if the height of the tower is 50 m.

Open in App

Solution

Let one of the persons. A be at a distance of $^{'} x^{'}$ m and the second person B be at a distance of $^{'} y^{'}$ m from the foot of the tower.
Given that angle of elevation of A is $30^{0}$
$t a n 30^{0} = 50 / x$
$1 / \sqrt{3} = 50 / x$
$x = 50 \sqrt{3} = 86.60 m$
And the angle of elevation of B is $38^{0}$
So. $t a n 38^{0} = 50 / y$
$0.7813 = 50 / y$
$y \approx 64 m$
Thus, the distance between A and B is x + y = 150.6 m

Suggest Corrections

Similar questions

Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be $30^{\circ}$ and $38^{\circ}$ respectively. Find the distance between them, if the height of the tower is 50 m. [ $tan 38^{\circ} = 0.7813$ ] [2 MARKS]

Q. Two men are on diametrically opposite side of a tower. They measure the angle of elevation of the top of the tower as

20^{0}

and

24^{0}

respectively. If the height of the tower is

40 m

, find the distance between them.
[Hint.use tables to find tan

20^{0}

and tan

24^{0}

]

From the top of tower $100 m$ height, a man observes two cars on the opposite side of tower with angles of depression $30^{\circ}$ and $45^{\circ}$ respectively. Find the distance between cars.

Join BYJU'S Learning Program