Question

1. solve for x. round to the nearest tenth of a degree, if necessary. 13. corey spots an airplane on radar that is currently approaching in a straight line, and that will fly directly overhead. the plane maintains a constant altitude of 6800 feet. corey initially measures an angle of elevation of 16° to the plane at point a. at some later time, he measures an angle of elevation of 40° to the plane at point b. find the distance the plane traveled from point a to point b. round your answer to the nearest foot if necessary. 12. hawa is flying a kite, holding her hands a distance of 3.75 feet above the ground and letting all the kites string out. she measures the angle of elevation from her hand to the kite to be 26°. if the string from the kite to her hand is 135 feet long, how many feet is the kite above the ground? round your answer to the nearest tenth of a foot if necessary.

Explanation:

Step1: Solve problem 11

We know that $\tan x=\frac{60}{63}$. So $x = \arctan(\frac{60}{63})$.
Using a calculator, $x=\arctan(\frac{60}{63})\approx43.6^{\circ}$ (rounded to the nearest tenth of a degree).

Step2: Solve problem 12

Let the height from the hand - level to the kite be $h$. We know that $\sin26^{\circ}=\frac{h}{135}$.
So $h = 135\times\sin26^{\circ}$.
$h=135\times0.4384\approx59.2$ feet.
The height of the kite above the ground is $h + 3.75=59.2+3.75 = 62.95\approx63.0$ feet (rounded to the nearest tenth of a foot).

Step3: Solve problem 13

Let the distance from the observer to the point directly below the plane at point $A$ be $d_1$ and at point $B$ be $d_2$.
We know that $\tan16^{\circ}=\frac{6800}{d_1}$, so $d_1=\frac{6800}{\tan16^{\circ}}=\frac{6800}{0.2867}\approx23718$ feet.
Also, $\tan40^{\circ}=\frac{6800}{d_2}$, so $d_2=\frac{6800}{\tan40^{\circ}}=\frac{6800}{0.8391}\approx8104$ feet.
The distance the plane traveled from point $A$ to point $B$ is $d_1 - d_2=23718 - 8104=15614$ feet (rounded to the nearest foot).

Answer:

1. $x\approx43.6^{\circ}$
2. $63.0$ feet
3. $15614$ feet