AB is a straight road leading to c the foot of a tower .A is at a distance of 200m from c and Bis125 m nearer .if the angle of elevation of the top of the tower at B is double the angle of elevation at A find the height of the tower
Answers
Answer:
The height of the tower is 100 meters.
Step-by-step explanation:
Let D be the top of tower,
height of tower = DC
length of base AC = 200 m
length of base BC = 200 - 125 = 75 m
let ∠A = Ф
=> ∠B = 2Ф
In triangle DCA,
tanФ = DC/AC = DC/200
=> DC = tanФ x 200...............eq1
similarly,
In triangle DCB,
tan2Ф = DC/BC = DC/75
=> DC = tan2Ф x 75...............eq2
from eqn1 and eqn2
tanФ x 200 = tan2Ф x 75
=> tanФ/tan2Ф = 75/200
=> tanФ/[2tanФ/1-tan²Ф] = 75/200
=> (1-tan²Ф)/2 = 75/200
=> 1-tan²Ф = 75/100
=> tan²Ф = 1 - 75/100 = 25/100
=> tanФ = 5/10
from eqn 1,
DC = tanФ x 200
=> DC = 5/10 x 200
=> DC = 100 meters.
Hence the height of the tower is 100 meters.
Hope it's helpful but didn't get the final answer. Try to check if any mistakes and please rectify them. Sorry for not giving the correct answer
