A boy is flying a kite with a string of length 100 m. If the string is tight and the angle of elevation of the kite is 26⁰ 32’, find the height of the kite correct to one decimal place, (ignore the height of the boy).

Consider AB as the height of kite A and AC as the string

The angle of elevation of the kite = 26⁰ 32’

Take AB = x m and AC = 100 m

We know that, sinθ = $\frac{\mathrm{AB}}{\mathrm{AC}}$

Substituting the values, we get:

=> sin 26⁰ 32’ = $\frac{\mathrm{x}}{100}$

=> 0.4387 = $\frac{\mathrm{x}}{100}$

=> x = 100 × 0.4387

=> x = 43.87 m

Hence, the height of the kite is 43.87 m.