The length of the string between a kite and a point on the ground, without any slack, is 102 m. If the string makes an angle with the level ground such that tan $$\alpha$$​ =$$\frac{15}8$$​, how high is the kite?

Correct option is B

Given:

The length of the string between a kite and a point on the ground, without any slack, is 102 m

​$$\tan \alpha = \frac{15}{8}$$ 

height of kite = ?

Concept Used: 

$$tan\alpha= \frac{perpendicular}{base}$$ 

Pythagorean theorem for right triangle: $$H^2 = P^2+B^2$$ ​

Solution: 

​$$\tan \alpha = \frac{h}{d} = \frac{15}{8} \quad \Rightarrow \quad h = \frac{15}{8} \cdot d$$ ​​

by the Pythagorean theorem:​

$$102^2 = \left( \frac{15}{8} \cdot d \right)^2 + d^2$$ 

$$10404 = \left( \frac{225}{64} \cdot d^2 \right) + d^2$$ 

$$10404 = \left( \frac{225}{64} + 1 \right) d^2$$ 

$$10404 = \left( \frac{225+64}{64} \right) d^2$$ 

$$10404 = \frac{289}{64} \times d^2$$

​​$$d^2 = \frac{10404 \times 64}{289}$$ 

$$d^2 = \frac{665856}{289} \approx 2304$$ 

$$d \approx \sqrt{2304} = 48 \, \text{m}$$ 

Now that we know the horizontal distance d=48 m, we can find the height h using the equation: 

​$$h = \frac{15}{8} \times d = \frac{15}{8} \times 48 = 90 \, \text{m}$$​​

Thus the height of the kite is 90 meters.