Step 1: Use the tangent ratio together with Pythagoras' theorem as simultaneous equations, instead of using sine and cosine separately.
Let $H$ be the height of the kite above the ground and $x_1$ be the horizontal distance from the first boy to the point on the ground directly below the kite.
Using the angle of elevation $30^\circ$ and the tangent ratio:
\[ \tan(30^\circ) = \frac{H}{x_1} \implies H = \frac{x_1}{\sqrt{3}} \quad \text{(Equation 1)} \]
The string itself is the hypotenuse of this right triangle, with length $60$ m, so by Pythagoras' theorem:
\[ x_1^2 + H^2 = 60^2 = 3600 \quad \text{(Equation 2)} \]
Step 2: Substitute Equation 1 into Equation 2 to solve for x1 first.
\[ x_1^2 + \left(\frac{x_1}{\sqrt{3}}\right)^2 = 3600 \]
\[ x_1^2 + \frac{x_1^2}{3} = 3600 \]
\[ \frac{3x_1^2 + x_1^2}{3} = 3600 \]
\[ \frac{4x_1^2}{3} = 3600 \]
\[ x_1^2 = 2700 \]
\[ x_1 = \sqrt{2700} = \sqrt{900 \times 3} = 30\sqrt{3} \text{ m} \]
Step 3: Substitute back into Equation 1 to find H.
\[ H = \frac{x_1}{\sqrt{3}} = \frac{30\sqrt{3}}{\sqrt{3}} = 30 \text{ m} \]
So the height of the kite above the ground is $30$ m.
Step 4: Handle the second boy's triangle, on the rooftop.
The second boy stands on a $20$ m building, so the kite's height above his eye level (roof) is:
\[ H - 20 = 30 - 20 = 10 \text{ m} \]
Let $x_2$ be his horizontal distance from the point below the kite. Using the $45^\circ$ angle of elevation:
\[ \tan(45^\circ) = \frac{10}{x_2} \implies 1 = \frac{10}{x_2} \implies x_2 = 10 \text{ m} \]
Step 5: Add the two horizontal distances, since the boys are on opposite sides of the kite.
\[ \text{Distance} = x_1 + x_2 = 30\sqrt{3} + 10 \]
Using $\sqrt{3} = 1.73$:
\[ \text{Distance} = 30(1.73) + 10 = 51.9 + 10 = 61.9 \text{ m} \]
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.
\[ \boxed{H = 30 \text{ m},\ \text{Distance} = 61.9 \text{ m}} \]