Suppose you are shooting an arrow from the top of a building at a height of $6 m$ to a target on the ground at an angle of depression of $60^{\circ}$ . What is the distance between you and the object?

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Solution

Let $A B$ be the height of the building.
Angle of depression $= 60^{\circ}$
$∠ C A B = 30^{\circ}$
In $△ A B C$ ,
$c o s 30^{\circ} = \frac{A B}{A C}$
or, $\frac{\sqrt{3}}{2} = \frac{6 m}{A C}$
or, $A C = \frac{12 m}{\sqrt{3}}$
or, $A C = 4 \sqrt{3} m = 6.93 m$
So, distance between you and the object $= 6.93 m$

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Suppose you are shooting an arrow from the top of a building at a height of 6 m to a target on the ground at an angle of depression of 60∘. What is the distance between you and the object?

Let AB be the height of the building.Angle of depression =60∘∠CAB=30∘In △ABC,cos30∘=ABACor, √32=6 mACor, AC=12 m√3or, AC=4√3 m=6.93 mSo, distance between you and the object =6.93 m

Suppose you are shooting an arrow from the top of a building at a height of $6 m$ to a target on the ground at an angle of depression of $60^{\circ}$ . What is the distance between you and the object?