Question

From a point on the ground, the angle of elevation of the top of tower is observed to be 60 dgree, from a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 45o. Find the height of the tower and its horizontal distance from the point of observation.​

Answer 1

Answer:

high frd tya hya hybrid hcd thu

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

From a point on the ground, the angle of elevation of the top of tower is observed to be 60°, and from a point 40 m vertically above the first point of observation, the angle of elevation of the top of the tower is 45°.

Let assume that

➢ AB represents the tower with base A and top B.

and

➢ C be any point on the ground, from where the angle of elevation of top of the tower is 60°.

➢ And D be point above C, 40 m vertically above, from where the angle of elevation of top of tower B is 45°.

Let assume that,

➢ BE = x meter

➢ AC = DE = y m

➢ and CD = AE = 40 meter.

Now,

$\red{\rm :\longmapsto\: In \: \triangle \: BED, \: we \: have}$

$\rm :\longmapsto\:tan45 \degree \: = \: \dfrac{BE}{ED}$

$\rm :\longmapsto\:1 = \dfrac{x}{y}$

$\bf\implies \:y = x - - - (1)$

Now, Consider

$\red{\rm :\longmapsto\: In \: \triangle \: ACB, \: we \: have}$

$\rm :\longmapsto\:tan60 \degree \: = \: \dfrac{AB}{AC}$

$\rm :\longmapsto\: \sqrt{3} = \dfrac{40 + x}{y}$

$\rm :\longmapsto\: \sqrt{3} = \dfrac{40 + x}{x} \: \: \: \: \: \: \: \: \{ \because \: x = y \}$

$\rm :\longmapsto\: \sqrt{3}x = 40 + x$

$\rm :\longmapsto\: \sqrt{3}x - x = 40$

$\rm :\longmapsto\: (\sqrt{3} -1) x = 40$

$\rm :\longmapsto\:x = \dfrac{40}{ \sqrt{3} - 1}$

On rationalizing the denominator, we get

$\rm :\longmapsto\:x = \dfrac{40}{ \sqrt{3} - 1} \times \dfrac{ \sqrt{3} + 1}{ \sqrt{3} + 1}$

$\rm :\longmapsto\:x = \dfrac{40( \sqrt{3} + 1)}{ {( \sqrt{3} )}^{2} - {1}^{2} }$

$\rm :\longmapsto\:x = \dfrac{40( \sqrt{3} + 1)}{ 3 - 1}$

$\rm :\longmapsto\:x = \dfrac{40( \sqrt{3} + 1)}{2}$

$\rm :\longmapsto\:x = 20( \sqrt{3} + 1)$

$\rm :\longmapsto\:x = 20( 1.732 + 1)$

$\rm :\longmapsto\:x = 20( 2.732)$

$\bf\implies \:x = 54.64 \: m$

Hence,

$\boxed{ \rm \:Height \: of \: tower, \: AB = 40 + x = 40 + 54.64 = 94.64 \: m}$

and

$\boxed{ \rm \:Horizontal \: distance, \: AC = y= x = 40 \: m}$