Answer:
Let PR be the tower and PQ be the ground.
Let point Q be at a distance of 30 m from the tower
Given 30
o
is the angle of elevation
∴tan30
∘
=
PQ
PR
⇒PR=PQtan30
⇒PR=
3
30
⇒PR=10
m
∴ Height of the tower is 10
m.
MARSHMELLOW FOLLOW ME
We have to find the height of the tower.
Let us consider the height of the tower as AB, the distance between the foot of the tower to the point on the ground as BC.
In ΔABC, trigonometric ratio involving AB, BC and ∠C is tan θ.
tan C = AB/BC
tan 30° = AB/30
1/√3 = AB/30
AB = 30/√3
= (30 × √3) / (√3 × √3)
= (30√3)/3
= 10√3
Height of tower AB = 10√3 m.
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Verified answer
Answer:
Let PR be the tower and PQ be the ground.
Let point Q be at a distance of 30 m from the tower
Given 30
o
is the angle of elevation
∴tan30
∘
=
PQ
PR
⇒PR=PQtan30
∘
⇒PR=
3
30
⇒PR=10
3
m
∴ Height of the tower is 10
3
m.
MARSHMELLOW FOLLOW ME
Answer:
We have to find the height of the tower.
Let us consider the height of the tower as AB, the distance between the foot of the tower to the point on the ground as BC.
In ΔABC, trigonometric ratio involving AB, BC and ∠C is tan θ.
tan C = AB/BC
tan 30° = AB/30
1/√3 = AB/30
AB = 30/√3
= (30 × √3) / (√3 × √3)
= (30√3)/3
= 10√3
Height of tower AB = 10√3 m.