Question

the angle of elevation of a flying kite is 77°4610. if the other end of the 198 - foot long string attached to the kite is tied to the ground, what is the approximate height of the kite? round your solution to the nearest foot.

Explanation:

Step1: Convert the angle to decimal degrees

We know that $1^{\circ}=60'$ and $1' = 60''$. So, $46'=\frac{46}{60}\approx0.7667^{\circ}$ and $10''=\frac{10}{3600}\approx0.0028^{\circ}$. Then $77^{\circ}46'10''\approx77 + 0.7667+0.0028=77.77^{\circ}$.

Step2: Use the sine - function

Let the height of the kite be $h$. We have a right - triangle where the length of the string is the hypotenuse $c = 198$ feet and the height of the kite is the side opposite the angle of elevation $\theta=77.77^{\circ}$. The sine of an angle in a right - triangle is defined as $\sin\theta=\frac{opposite}{hypotenuse}$. So, $\sin(77.77^{\circ})=\frac{h}{198}$.

Step3: Solve for $h$

We can rewrite the equation as $h = 198\times\sin(77.77^{\circ})$. Since $\sin(77.77^{\circ})\approx0.977$, then $h=198\times0.977 = 193.446\approx193$ feet.

Answer:

193