Question

The vertical angle to the top of a flagpole from point A on the ground is observed to be 37°11'. The observer walks 17 m directly away from point A and the flagpole to point B and finds the new angle to be 25°43'. What is the approximate height of the flagpole?

Answer 1

22m

Explanation:

Let height of flagpole=h

AB==17 m

`\angle CAD=37^(\circ)11'=37+(11)/(60)=37.183^(\circ)`(1 degree= 60 minute)

`\angle B=25^(\circ)43'=25+(43)/(60)=25.72^(\circ)`

We have to find the approximate height of the flagpole.

In triangle CDA,

`(CD)/(DA)=tan\theta=(Perpendicular\;side)/(Base)`

`(h)/(DA)=tan37.183^(\circ)`

`h=DA(0.759)`

In triangle CDB,

`tan 25.72^(\circ)=(CD)/(DB)`

`0.482=(h)/(DA+17)`

`0.482DA+8.194=h`

Substitute the value

`0.482DA+8.194=0.759DA`

`8.194=0.759DA-0.482DA`

`8.194=0.277DA`

`DA=(8.194)/(0.277)=29.58`

Substitute the value

`h=29.58 * 0.759=22.45 m\approx 22m`

Hence, the height of the flagpole=22 m