Question

A boy standing on a horizontal plane is flying a kite with a string of length 60 m, at an angle of elevation of $30^\circ$. Another boy standing on the roof of a 20 m high building, finds the angle of elevation of same kite to be $45^\circ$. If both the boys are on opposite sides of the kite, find the distance of the first boy from the base of the building. Also, find the height of the kite from the ground. (Use $\sqrt{3} = 1.73$)

Hint:

Always pay attention to the phrase "opposite sides of the kite".
This means you must ADD the two horizontal distances ($D = x_1 + x_2$).
If they were on the "same side", you would have to SUBTRACT them ($D = x_1 - x_2$).

Answer 1

Step 1: Understanding the Question:
This question is a application of trigonometry (Heights and Distances).
We have a kite flying at some height above the ground.
Two boys are flying/observing the kite from opposite sides:
- The first boy is on the ground, flying the kite with a $60$ m string at an elevation of $30^\circ$.
- The second boy is on the roof of a $20$ m tall building, observing the same kite at an angle of elevation of $45^\circ$.
We need to find the total horizontal distance between the first boy and the building, as well as the height of the kite from the ground.

Step 2: Key Formula or Approach:
1. Model the scenario with right-angled triangles.
2. Use the sine trigonometric ratio in the first boy's triangle to find the height of the kite.
3. Use the tangent trigonometric ratio in the second boy's triangle to find his horizontal distance from the kite.
4. The total distance from the first boy to the building is the sum of their individual horizontal distances to the kite.

Step 3: Detailed Explanation:

• Height of the kite from the ground ($H$):
Let the position of the kite be $K$, and the first boy be $B_1$ on the ground.
In right-angled triangle $\Delta K B_1 P$ (where $P$ is the point on the ground directly beneath the kite):
- Hypotenuse (string length) $= 60\text{ m}$
- Angle of elevation $= 30^\circ$
Apply the sine ratio:
\[ \sin(30^\circ) = \frac{\text{Height of kite } (H)}{\text{String length}} \] \[ \frac{1}{2} = \frac{H}{60} \implies H = 30\text{ m} \] So the height of the kite from the ground is $30$ m.

• Horizontal distance of first boy to the kite ($x_1$):
Using the cosine ratio in the same triangle:
\[ \cos(30^\circ) = \frac{x_1}{60} \] \[ \frac{\sqrt{3}}{2} = \frac{x_1}{60} \implies x_1 = 30\sqrt{3}\text{ m} \]

• Horizontal distance of building to the kite ($x_2$):
Let the second boy be $B_2$ on the roof of a building of height $20$ m.
The height of the kite above the roof level is:
\[ h' = H - 20 = 30 - 20 = 10\text{ m} \] In the right-angled triangle formed at the roof level, the angle of elevation is $45^\circ$.
Apply the tangent ratio:
\[ \tan(45^\circ) = \frac{h'}{x_2} \] \[ 1 = \frac{10}{x_2} \implies x_2 = 10\text{ m} \]

• Total distance between the first boy and the building ($D$):
Since the boys are on opposite sides of the kite, the total horizontal distance is the sum of $x_1$ and $x_2$:
\[ D = x_1 + x_2 = 30\sqrt{3} + 10 \] Substitute $\sqrt{3} = 1.73$:
\[ D = 30(1.73) + 10 = 51.9 + 10 = 61.9\text{ m} \]

Step 4: Final Answer:
The height of the kite from the ground is $30$ m, and the distance of the first boy from the base of the building is $61.9$ m.