From the top of a cliff $92 m$ high, the angle of depression of a buoy is $20^{o}$ . Calculate, to the nearest, a distance of the buoy from the foot of the cliff.

Open in App

Solution

Let $A B$ be the cliff and $C$ be the buoy.
$A B = 92 m (Given)$
Also,
$∠ A C B = 20 °$
Therefore,
In $△ A B C$ ,
$tan 20 ° = \frac{A B}{B c}$
$\Rightarrow B C = \frac{92}{0.364} \approx 253 m$
Hence the buoy is at a distance of $253 m$ from the foot of the cliff.

Suggest Corrections

Similar questions

From the top of a cliff of $92 m$ height, the angle of depression of a buoy is $20^{\circ}$ . Calculate to the nearest meter, the distance of the buoy from the foot of the cliff. {Use $tan 20^{o} = 0.36$ }

Q. From the top of a cliff, the angle of depression of car on the ground is

60^{o}

. If the car is at a distance of 30 m from the cliff. Find the height of the cliff.

Q. A man 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^{o}

and the angle of depression of its base is

30^{o}

. Calculate the distance of the cliff from the ship and the height of the cliff.

Join BYJU'S Learning Program

From the top of a cliff 92 m high, the angle of depression of a buoy is 20o. Calculate, to the nearest, a distance of the buoy from the foot of the cliff.

Let AB be the cliff and C be the buoy.AB=92m(Given)Also,∠ACB=20°Therefore,In △ABC,tan20°=ABBc⇒BC=920.364≈253mHence the buoy is at a distance of 253m from the foot of the cliff.

From the top of a cliff $92 m$ high, the angle of depression of a buoy is $20^{o}$ . Calculate, to the nearest, a distance of the buoy from the foot of the cliff.

Let $A B$ be the cliff and $C$ be the buoy.
$A B = 92 m (Given)$
Also,
$∠ A C B = 20 °$
Therefore,
In $△ A B C$ ,
$tan 20 ° = \frac{A B}{B c}$
$\Rightarrow B C = \frac{92}{0.364} \approx 253 m$
Hence the buoy is at a distance of $253 m$ from the foot of the cliff.