Question

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Answer 1

Given AB is a straight road leading to a foot of tower at C. A being at the distance of 200m from C and B is 125m nearer. Angle of elevation of the top of tower at B is double than at A. So find the height of the tower.

Let the height of the tower be h, and it will be DC. Join AD and BD. Let angle A = x and angle B = 2x

tan x = h/200 or h = 200 tan x------------(1)

tan 2x = h/75 or h = 75 tan2 x

tan 2x = (2 tan x/1 - tan²x)

h = 75(2 tan x / 1 - tan²x)-----------------(2)

From (1) and (2) we get

200 tan x = 75(2 tan x /1 - tan²x)

1 - tan²x = 3/4

tan²x = 1/4

tan x = 1/2------------(3)

Substituting (3) in (1) we get

h = 200 x 1/2

h = 100 m

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Answer 2

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° = h45

=> h = 45 tan 2x  => h = 45 2tanx 1-tan2x   .............. (1) tan 2x = 2tan x1-tan2x

In Triangle, ACD

tan x = h120

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

From (1) and (2)

120 tanx = 45×2 tanx 1-tan2x

=> 1-tan2x = 34

=> tan2x = 14

=> tan x = 12

Put value of tan x in (2)

we get h = 60 

so height of tower is 60cm

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