Question

angle of elevation of the top of a tower is 45 degree from a point. on the ground on walking 1 km towards the tower, angle is found to be 60 calculate the height of the tower

Answer 1

Step-by-step explanation:

Given that:Angle of elevation of the top of a tower is 45 degree from a point. on the ground on walking 1 km towards the tower, angle is found to be 60.

To find:

calculate the height of the tower

Solution: Draw the figure from the given situation

Let the height of tower is AB

In ∆ABC

$tan \: 45° = \frac{AB}{AC} \\ \\1 = \frac{AB}{x + 1} \\ \\ AB = x + 1 \: \: \: ...eq1$

In ∆ADB

$tan \: 60° = \frac{AB}{AD} \\ \\ \sqrt{3} = \frac{AB}{x} \\ \\ x= \frac{AB}{ \sqrt{3} } \: \: \: ...eq2$

Put the value of x from eq2 in eq1

$AB = \frac{AB}{ \sqrt{3} } + 1 \\ \\ AB(1 - \frac{1}{ \sqrt{3} } ) = 1 \\ \\ AB = \frac{ \sqrt{3} }{ \sqrt{3} - 1 }$

Rationalize the denominator

$AB = \frac{ \sqrt{3} }{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } \\ \\ AB = \frac{3 + \sqrt{3} }{3 - 1} \\ \\ AB = \frac{3 + \sqrt{3} }{2} \: km$

Height of tower is (3+√3) /2 km.

Hope it helps you.

Answer 2

Answer:

Height of tower is (3+√3) /2 km.