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AB is a straight road leading …

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AB is a straight road leading to C, the foot of a tower, A being at a distance of 120m from C and B being 75m nearer. If the angle of elevation of the tower at B is the double of the angle of elevation of the tower at A, find the height of the tower.

Posted by Shruti Sharma 7 years, 3 months ago

CBSE > Class 10 > Mathematics

2 answers

Naveen Sharma 7 years, 3 months ago

Just consider the image for Reference.

Gievn : AB is straigt Road and C is the foot of tower. Let DC is tower of height h, and AC is 120cm and BC is 45 cm as it is 75cm nearer than A.

Solution : In Triangle BCD

$\tan 60 ° = \frac{h}{45}$

=> $h = 45 \tan 2 x = > h = 45 \frac{2 \tan x}{1 - \tan^{2} x} . . . . . . . . . . . . . . (1) \tan 2 x = \frac{2 \tan x}{1 - \tan^{2} x}$

In Triangle, ACD

$\tan x = \frac{h}{120}$

=> $h = 120 \tan x . . . . . . . . . . . . . . . . . . . . (2)$

From (1) and (2)

$120 \tan x = \frac{45 \times 2 \tan x}{1 - \tan^{2} x}$

=> $1 - \tan^{2} x = \frac{3}{4}$

=> $\tan^{2} x = \frac{1}{4}$

=> $\tan x = \frac{1}{2}$

Put value of tan x in (2)

we get h = 60

so height of tower is 60cm

Shweta Gulati 7 years, 3 months ago

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