Question

A kite is being flown at a 45 angle. The string of the kite is 120 feet long. How high is the kite above the point at which the string is held?

Answer 1

Answer: The height of the kite above the point at which the string is held is $120\sqrt{2}$ feet.

Step-by-step explanation:

Given : A kite is being flown at $45^{\circ}$ . The string of the kite is 120 feet long.  

Let AB denote the string of kite and AC be the height of the kite above the point at which the string is held.

Now, in right Δ ABC

$\sin45^{\circ}=\frac{AC}{AB}\\\Rightarrow\ \frac{1}{\sqrt{2}}=\frac{AC}{120}\\\Rightarrow\ AC=120\sqrt{2}$

hence, The height of the kite above the point at which the string is held is $120\sqrt{2}$ feet.

Answer 2

Answer:

height will be =84.86 feet.

Step-by-step explanation:

It is given that A kite is being flown at a 45° angle which means angle of elevation is 45°. Also, the length of the string of the kite=120 feet.

Thus,

$\frac{AB}{AC}=sin45^{\circ}$

⇒$\frac{AB}{120}=\frac{1}{\sqrt{2}}$

⇒$AB=120{\times}\frac{1}{\sqrt{2}}$

⇒$AB=\frac{120}{1.414}$

⇒$AB=84.86 feet$

Thus, the height will be =84.86 feet.