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

Answer:

22m

Step-by-step explanation:

Let height of flagpole=h

AB==17 m

$\angle CAD=37^{\circ}11'=37+\frac{11}{60}=37.183^{\circ}$(1 degree= 60 minute)

$\angle B=25^{\circ}43'=25+\frac{43}{60}=25.72^{\circ}$

We have to find the approximate height of the flagpole.

In triangle CDA,

$\frac{CD}{DA}=tan\theta=\frac{Perpendicular\;side}{Base}$

$\frac{h}{DA}=tan37.183^{\circ}$

$h=DA(0.759)$

In triangle CDB,

$tan 25.72^{\circ}=\frac{CD}{DB}$

$0.482=\frac{h}{DA+17}$

$0.482DA+8.194=h$

Substitute the value

$0.482DA+8.194=0.759DA$

$8.194=0.759DA-0.482DA$

$8.194=0.277DA$

$DA=\frac{8.194}{0.277}=29.58$

Substitute the value

$h=29.58 \times 0.759=22.45 m\approx 22m$

Hence, the height of the flagpole=22 m