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 \(30^\circ\). From the bottom of the same building, the angle of elevation of kite is \(45^\circ\). 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.