Question

A girl standing on a cliff sees a boat at an angle of depression of
45*. If the cliff is 30m tall, how far from the cliff is the boat?​

Answer 1

Given:-

• Angle of depression = ∠ABC = 45°
• Height of cliff = AB = 30m

To Find:-

• Distance between the cliff and the boat (BC)

Identities:-

$\sf tan45 \degree = \dfrac{Perpendicular}{Base} = \dfrac{AB}{BC}$

Solution :-

$\sf \implies \: tan45 \degree = \dfrac{AB}{BC}$

$\sf \implies \: 1 = \dfrac{30}{BC}$

$\sf \implies \boxed{\boxed{ \purple{\bf{BC = 30m}}}}$

$\bf \therefore \underline{distance \: between \: cliff \: and \: boat \: is \: 30m}.$

__________________

Important identities of trigonometry:-

$\huge\displaystyle\begin{tabular}{|c|c|c|c|c|c|}\cline{1-6}&0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}\cline{1-6}\cline{1-6}Sin&0& \dfrac{1}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{\sqrt{3}}{2} &1\cline{1-6}\cline{1-6}Cos&1& \dfrac{\sqrt{3}}{2} & \dfrac{1}{\sqrt{2}} & \dfrac{1}{2} &0\cline{1-6}\cline{1-6}Tan&\infty& \dfrac{1}{\sqrt{3}} &1& \sqrt{3} &0 \cline{1 - 5}\cline{1 - 5}\end{tabular}$

Angle of depression:-

When a observer is higher than the object they are looking at, in this condition angle of depression is formed.

Whereas angle of elevation form up when observer stands on the ground and look at the top of object which is bigger in length.

Answer 2

Given ,

Angle of depression (∠ABC) = 45

Height of cliff (AC) = 30

We know that ,

$\boxed{ \sf{Tan \: ( \theta) = \frac{Perpendicular}{Base} }}$

Thus ,

Tan(45) = 30/BC

1 = 30/BC

BC = 30/1

BC = 30 m

$\therefore$ The distance between cliff and boat is 30 m