Question

A kite is flying at a height of 60 m above the ground level. Ravi, standing at the roof of the house is holding the string straight and observes the angle of elevation of kite as 30º. From the bottom of the same building, the angle of elevation of kite is 45º. Find the length of the string and height of roof from the ground. (Use \(\sqrt{3} = 1.73\))

Hint:

An angle of elevation of \(45^\circ\) always forms an isosceles right-angled triangle.
This immediately tells you that the horizontal ground distance is equal to the vertical height, which is 60 m!

Answer 1

Step 1: Find the horizontal distance from the bottom of the building.
From the bottom, the angle of elevation is $45^\circ$, so $\tan45^\circ=\frac{60}{x} \implies x=60$ m, since $\tan45^\circ=1$.
Step 2: Relate the height above the roof to $x$ using tangent.
Let $y=60-h$ be the vertical rise of the kite above roof level. From the roof, $\tan30^\circ=\frac{y}{x} \implies y=60\times\frac{1}{\sqrt3}=\frac{60}{\sqrt3}=20\sqrt3$ m (after rationalizing).
Step 3: Find the height of the roof.
$h=60-y=60-20\sqrt3=60-20(1.73)=60-34.6=25.4$ m.
Step 4: Find the string length using sine.
$\sin30^\circ=\frac{y}{L} \implies L=\frac{y}{\sin30^\circ}=\frac{20\sqrt3}{\frac12}=40\sqrt3=40\times1.73=69.2$ m.
\[ \boxed{L=69.2\text{ m},\ h=25.4\text{ m}} \]