One flies a kite with a thread 150 metre long. If the thread of the kite makes an angle of 60° with the horizontal line, then the height of the kite from the ground (assuming the thread to be in a straight line) is

AB = Length of the thread = 150 metre ∠BAC = 60° In $riangle A B C, sin 60^{circ}=frac{B C}{A B} Rightarrow frac{sqrt{3}}{2}=frac{B C}{150}$ $Rightarrow B C=150 imes frac{sqrt{3}}{2}=75 sqrt{3}$ metre Alternative:

$an 60^{circ}=frac{sqrt{3}}{1}=frac{mathrm{AB}}{mathrm{BC}}$

$mathrm{AC}=sqrt{(sqrt{3})^{2}+(1)^{2}}=2$

According to the question,

2 unit = 150 metre

1 unit $=frac{150}{2}$

$sqrt{3}$ unit $=frac{150}{2} imes sqrt{3}=75 sqrt{3}$ metre

Height of kite from the ground $=75 sqrt{3}$ metre