Question

from the top of the building 12m high,,the angle of elevation of the top of a tower is found to be 60°and the angle of depression of the base of the tower is 45° find the height of the tower and it's distance on the ground from the building

Answer 1

easy i know

answer is 12root3

let x be distance on the ground form the building and let y be the height of tower

tan45=P/b

1=12/x

then x=12m

now

tan60=p/b

root3=y/x

root3x=y

root3 x 12=y

y=12root3

done

Answer 2

Height of tower is 12(1+√3) and it's distance on the ground from building is 12m.

• Let building of height 12m as AB and tower of x height as CE

then according to the question data ,the angle of elevation of the top of a tower is found to be 60°and the angle of depression of the base of the tower is 45°.

• Let draw a line parallel to ground from top of the building intersecting tower at a height of CD = 12m. Name intersecting point as D.
• So, angle ACB will also equal to 45°. (alternate interior angle)
• We have to find value of BC and CE.
• Using tanθ in triangle ABC, we have

tan45° = AB/BC = 12/BC

BC = 12m = AD

• Using tanθ in triangle ADE, we have

tan60° = DE/AD = DE/12

DE = 12√3m

• Height of tower is CE = CD+DE = 12+12√3 = 12(1+√3)m
• Hence, height of tower is 12(1+√3) and it's distance on the ground from building is 12m.