Step 1: Let $K$ be the kite at height $h$ above the ground.
For the first boy, string $= 60$ m and angle of elevation $= 30^\circ$.
$$\sin 30^\circ = \frac{h}{60} \Rightarrow \frac{1}{2} = \frac{h}{60}$$
Step 2: $$h = 30\ \text{m}$$
Hence the height of the kite from the ground is $30$ m.
Step 3: Horizontal distance of the first boy from the point below the kite :
$$\cos 30^\circ = \frac{d_1}{60} \Rightarrow d_1 = 60 \times \frac{\sqrt{3}}{2} = 30\sqrt{3}$$
$$d_1 = 30 \times 1{\cdot}73 = 51{\cdot}9\ \text{m}$$
Step 4: For the second boy on the $20$ m roof, the vertical rise up to the kite is
$$30 - 20 = 10\ \text{m}$$
Step 5: With angle of elevation $45^\circ$, the horizontal distance $d_2$ satisfies
$$\tan 45^\circ = \frac{10}{d_2} \Rightarrow 1 = \frac{10}{d_2} \Rightarrow d_2 = 10\ \text{m}$$
Step 6: Since the boys are on opposite sides of the kite, the distance of the first boy from the base of the building is
$$d_1 + d_2 = 51{\cdot}9 + 10 = 61{\cdot}9\ \text{m}$$
Hence the height of the kite $= 30$ m and the required distance $= 61{\cdot}9$ m.