Question

A man standing on the deck of a ship is 12 m above water level.He observes that the angle of elevation of the top of a cliff is 45 degree and the angle of depression of the base of the cliff is 30 degree.calculate the distance of the cliff from the ship and the height of the cliff.(take √3=1.732)

Answer 1

Hope this will help you!

Answer 2

Answer:

The distance of the cliff from the ship is 20.784 m and Height of cliff is 32.784 m .

Step-by-step explanation:

Refer the attached figure.

A man standing on the deck of a ship is 12 m above water level.i.e.BE = DC=12 m

The angle of elevation of the top of a cliff is 45 degree i.e. ∠ABC=45°

The angle of depression of the base of the cliff is 30 degree i.e. ∠DBC=45°

Height of the cliff = AD

The distance of the cliff from the ship =BC

Using trigonometric ratio

In ΔABC

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 45^{\circ}= \frac{AC}{BC}$

$1= \frac{AC}{BC}$

$AC=BC$

In ΔBCD

$Tan \theta = \frac{Perpendicular}{Base}$

$Tan 30^{\circ}= \frac{DC}{BC}$

$\frac{1}{\sqrt{3}}= \frac{12}{BC}$

$BC=12\sqrt{3}$

$BC=12\times 1.732$

$BC=20.784$

The distance of the cliff from the ship =BC=20.784 m

BC =AC = 20.784 m

Height of cliff = AD = AC+DC= 20.784+12=32.784 m

Hence The distance of the cliff from the ship is 20.784 m and Height of cliff is 32.784 m .