Answer:
22m
Step-by-step explanation:
Let height of flagpole=h
AB==17 m
[tex]\angle CAD=37^{\circ}11'=37+\frac{11}{60}=37.183^{\circ}[/tex](1 degree= 60 minute)
[tex]\angle B=25^{\circ}43'=25+\frac{43}{60}=25.72^{\circ}[/tex]
We have to find the approximate height of the flagpole.
In triangle CDA,
[tex]\frac{CD}{DA}=tan\theta=\frac{Perpendicular\;side}{Base}[/tex]
[tex]\frac{h}{DA}=tan37.183^{\circ}[/tex]
[tex]h=DA(0.759)[/tex]
In triangle CDB,
[tex]tan 25.72^{\circ}=\frac{CD}{DB}[/tex]
[tex]0.482=\frac{h}{DA+17}[/tex]
[tex]0.482DA+8.194=h[/tex]
Substitute the value
[tex]0.482DA+8.194=0.759DA[/tex]
[tex]8.194=0.759DA-0.482DA[/tex]
[tex]8.194=0.277DA[/tex]
[tex]DA=\frac{8.194}{0.277}=29.58[/tex]
Substitute the value
[tex]h=29.58 \times 0.759=22.45 m\approx 22m[/tex]
Hence, the height of the flagpole=22 m
