Find the angle of elevation of a point on the ground to the top of a tower if the horizontal distance between them is √3 times the height of tower.
Answers
Answered by
0
Answer:
=34.60m
Step-by-step explanation:
To find → Height of tower (AB)
Let AB =h m
In △ABC, by Trigonometry
tan60
∘
=
BC
AB
3
=
BC
h
⇒BC=(
3
h
)m
Since BCDE is a ||gm, so BC=DE and CD = BE
∴DE=(
3
h
)m;BE=40m
Now in △ADE
tan30
∘
=
DE
AE
3
1
=
3
h
h−40
h=3h−120
2h=120m
h=60m
Height of tower = 60 m.
To find → Horizontal Distance from point of observation (BC)
BC=
3
h
=
3
60
=20
3
m
BC=20
3
m=34.60m
Answered by
1
Step-by-step explanation:
Hence, the angle of elevation is 30°
Attachments:
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