The angle of elevation of the top of a tower from two points at distances of 5 mts & 15 mts from
the base of the tower and in the same straight line with it are complementary. Find the height of the
tower. (
√
3 = 1.732)
Answers
Answer:
Height of the tower = 8.66 mts
Step-by-step explanation:
Let the height of the tower = H
Angle of elevation for point at a distance 5 m = a°
Angle of elevation for point at a distance 15 m = b°
As per the condition given, both angles of elevation are complementary.
i.e.
a° + b° = 90° ......(1)
Consider side of traingle with distance 5m.
We can write,
Tan a° = Height of tower ÷ Distance
∴ Tan a° = H / 5 .....(2)
Consider side of triangle with distance 15m
∴ Tan b° = Height of tower ÷ Distance
∴ Tan b° = H / 15 ..... (3)
But from eqn 1 ..... b° = (90 - a)°
Substituting in enq 3, we get
Tan (90-a)° = H / 15
since, Tan (90 - ∅) = Cot ∅ = 1 / Tan∅
we can write,
Tan (90-a)° = 1 / Tan a° = H / 15
∴ H × Tan a° = 15
(putting value of Tan a° from eqn 2...)
∴ H × (H / 5) = 15
∴ H² = 15 ×5
∴ H² = 75
∴ H = √75
∴ H = 5√3
(√3 = 1.732..)
∴ H = 5 × 1.732
∴ H = 8.66 mts
Height of the tower = 8.66 mts
Answer:
8.66 m
Step-by-step explanation:
Let say height of tower = H m
Angle of elevation from 5 meter = θ
then angle of elevation from 15 m = 90° - θ
Tan θ = H/5
Tan (90 - θ) = H/15
Tan (90 - θ) = Cotθ = 1/Tanθ = H/15
=> Tanθ = 15/H
Equating both
H/5 = 15/H
=> H² = 75
=> H = √75
=> H = 5√3
=> H = 5 * 1.732
=> H = 8.66 m