Question

The horizontal distance between two towers is 90 m. The angle of depression of the top of first tower. when seen from second tower is 30°. If the height of the second tower is 160 m, find the height of the first tower.

Answer 1

Answer:

Height of the first tower = 160 - 30√3 m

Step-by-step explanation:

Hi,

Let AD be the second tower, AD = 160 m

Let BE be the first tower,

Given the distance between 2 towers, AB = 90 m.

Given Angle of depression, ∠XDE = 30°

But ∠XDE = ∠CED

∠CED = 30°

In Δ CDE, tan ∠CED = CD/CE

tan 30° = CD/AB

⇒ CD = 90/√3 = 30√3.

But AD = AC + CD,

AD = BE + CD

BE = AD - CD

BE = 160 - 30√3

Hence, height of the first tower = 160 - 30√3 m .

Hope, it helps !

Answer 2

Answer:

108.04 m

Step-by-step explanation:

Define x:

Let the difference in the height of the two tower be x

Height of the first tower = 160 - x

Find x:

tan θ = opp/adj

tan(30º) = x/90

x = 90 x tan(30)

x = 51.96m

Find the height of the first tower:

Height = 160 - x

Height = 160 - 51.96 = 108.04 m

Answer: 108.04 m