From the top of a building $50 \sqrt{3} m$ high the angle of depression of a car on the ground is observed to be $60^{o}$ . Find the distance of the car from the building.

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Solution

Let $Q R$ be the distance of the car from the building.
Now, In $△ P Q R$
$\Rightarrow tan 60 ° = \frac{P Q}{Q R} = \frac{50 \sqrt{3}}{Q R}$
$\Rightarrow \sqrt{3} = \frac{50 \sqrt{3}}{Q R}$
$\Rightarrow Q R = 50 m$
$∴$ The car is at $50 m$ distant from the building.
Hence, the answer is $50 m .$

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From the top of a building 50√3m high the angle of depression of a car on the ground is observed to be 60o. Find the distance of the car from the building.

Let QR be the distance of the car from the building. Now, In △PQR⇒tan60°=PQQR=50√3QR⇒√3=50√3QR⇒QR=50m∴ The car is at 50m distant from the building. Hence, the answer is 50m.

From the top of a building $50 \sqrt{3} m$ high the angle of depression of a car on the ground is observed to be $60^{o}$ . Find the distance of the car from the building.