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: Understanding the Question:
The topic of this question is Some Applications of Trigonometry.
We are given a double-angle elevation scenario involving a kite \(K\) flying at a height of 60 m.
A building of height \(h\) stands at some horizontal distance from the line directly below the kite.
We are given two observer positions: the roof \(R\) of the building (angle of elevation is \(30^\circ\)) and the bottom \(B\) of the building (angle of elevation is \(45^\circ\)).
We need to find the length of the string held by Ravi from the roof and the height \(h\) of the building.

Step 2: Key Formula or Approach:
Let the horizontal distance from the building to the point directly below the kite be \(x\).
- From the bottom \(B\), we use the tangent ratio:
\[ \tan 45^\circ = \frac{\text{Opposite}}{\text{Adjacent}} \] - From the roof \(R\), the vertical height of the kite above the roof level is \(60 - h\). We use the tangent ratio:
\[ \tan 30^\circ = \frac{60 - h}{x} \] - To find the length of the string \(L\), we use the sine ratio from the roof level:
\[ \sin 30^\circ = \frac{60 - h}{L} \]

Step 3: Detailed Explanation:

• Define the variables:
Let the height of the building be \(h\) meters.
Let the ground distance to the point directly below the kite be \(x\) meters.

• Apply trigonometry in the right-angled triangle from the bottom \(B\):
\[ \tan 45^\circ = \frac{60}{x} \] Since \(\tan 45^\circ = 1\):
\[ 1 = \frac{60}{x} \implies x = 60 \text{ m} \]

• Apply trigonometry in the right-angled triangle from the roof \(R\):
The horizontal distance is still \(x = 60\) m, and the vertical height above the roof is \(60 - h\):
\[ \tan 30^\circ = \frac{60 - h}{60} \] Since \(\tan 30^\circ = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \frac{60 - h}{60} \] \[ 60 - h = \frac{60}{\sqrt{3}} \] Rationalize the fraction:
\[ 60 - h = 20\sqrt{3} \] \[ h = 60 - 20\sqrt{3} \]

• Calculate the numerical value of the building height \(h\) using \(\sqrt{3} = 1.73\):
\[ h = 60 - 20(1.73) \] \[ h = 60 - 34.6 = 25.4 \text{ m} \]

• Calculate the length of the string \(L\) from the roof \(R\):
Using the sine ratio in the right-angled triangle above the roof level:
\[ \sin 30^\circ = \frac{60 - h}{L} \] Since \(\sin 30^\circ = \frac{1}{2}\) and \(60 - h = 20\sqrt{3}\):
\[ \frac{1}{2} = \frac{20\sqrt{3}}{L} \] \[ L = 40\sqrt{3} \text{ m} \]

• Convert to a decimal value using \(\sqrt{3} = 1.73\):
\[ L = 40 \times 1.73 = 69.2 \text{ m} \]


Step 4: Final Answer:
The length of the string is 69.2 m, and the height of the roof from the ground is 25.4 m.