Question

From the top of a tower 100m high, a man observes two cars on the opposite sides of the tower with angles of depressions 30° and 45° respectively. Find the distance between the cars. *​

Answer 1

Given :-

• From the top of a tower 100m high, a man observes two cars on the opposite sides of the tower with angles of depressions 30° and 45° respectively.

To find :-

• The distance between the cars.

Solution :-

It is given that man observes two cars on the opposite sides of the tower

• Height of tower (AB) = 100m

• Angles of depression 30° and 45°

• 30° = ∠BCA

• 45° = ∠BDA

Let the CD be the distance between two cars

• In ∆BCA

→ tan 30° = BA/AC

→ 1/√3 = 100/AC

→ AC = 100√3 m

• In ∆BDA

→ tan 45° = BA/AD

→ 1 = 100/AD

→ AD = 100m

Now, distance between two cars

→ CD = AC + AD

Put the values of AC and AC

→ CD = 100√3 + 100

Take 100 as a common

→ CD = 100(√3 + 1)

Put the value of √3 = 1.73

→ CD = 100(1.73 + 1)

→ CD = 100 × 2.73

→ CD = 273 m

Hence,

• Distance between two cars is 273 m

Extra Information :-

• tan θ = perpendicular/base

• sin θ = perpendicular/hypotenuse

• cos θ = perpendicular/hypotenuse

• cot θ = base/perpendicular

• cosec θ = hypotenuse/perpendicular

• sec θ = hypotenuse/base