Question

the angle of depression of boat from the mid-point of vertical cliff is 45° if boat is 140 metres from the base of the cliff

Explanation:

Step1: Understand the problem

We have a vertical cliff, the mid - point of the cliff, and a boat. The angle of depression from the mid - point of the cliff to the boat is \(45^{\circ}\), and the horizontal distance between the boat and the base of the cliff is 140 meters. We need to find the height of the cliff.

Let the height of the cliff be \(h\) meters. Then the height from the mid - point of the cliff to the base of the cliff is \(\frac{h}{2}\) meters.

The angle of depression is equal to the angle of elevation from the boat to the mid - point of the cliff (alternate interior angles). So, the angle of elevation \(\theta = 45^{\circ}\).

The horizontal distance between the boat and the base of the cliff (adjacent side) \(x = 140\) meters, and the height from the mid - point of the cliff to the base of the cliff (opposite side) is \(\frac{h}{2}\) meters.

Step2: Use the tangent function

We know that in a right - triangle, \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)

For \(\theta = 45^{\circ}\), \(\tan(45^{\circ}) = 1\)

\(\tan(45^{\circ})=\frac{\frac{h}{2}}{140}\)

Since \(\tan(45^{\circ}) = 1\), we have \(1=\frac{\frac{h}{2}}{140}\)

Step3: Solve for \(h\)

From \(1=\frac{\frac{h}{2}}{140}\), we can cross - multiply:

\(\frac{h}{2}=1\times140\)

\(\frac{h}{2} = 140\)

Multiply both sides by 2: \(h=140\times2 = 280\) meters

Answer:

The height of the vertical cliff is 280 meters.