Question

From the top of the hill , the angles of depression of two consecutive kilometer stones due east are found to be 30 and 45 degrees, find the height of the hill

Answer 1

The height of hill is 1.366 km

Solution:

The figure is attached below

Let AB is the height of the hill and two stones are C and D respectively where depression is 45 degree and 30 degree

The distance between C and D is 1 km

CD = 1 km

Here depression and hill has formed right angle triangles with the base

To find: height of hill

height of hill = AB

In triangle ABC,

$tan 45 = \frac{height}{base}$

$tan 45 = \frac{AB}{BC}$

We know tan 45 (in degrees) = 1

$1 = \frac{AB}{BC}$

AB = BC ----- eqn 1

In triangle ABD,

$tan 30 = \frac{AB}{BD}$

From attached figure, BD = BC + CD

Also we know that,

$tan 30 = \frac{1}{\sqrt{3}} = \frac{1}{1.732}$

$\frac{1}{1.732} = \frac{AB}{BC + CD}$

As AB = BC from eqn 1 and CD = 1 km,

$\frac{1}{1.732} = \frac{AB}{AB + 1}$

$1.732AB = AB + 1$

$1.732AB -AB = 1\\\\AB( 1.732 - 1) = 1\\\\AB(0.732) = 1\\\\AB = \frac{1}{0.732}\\\\AB = 1.366$

Hence the height of hill is 1.366 km