Question

write the trigonometric equation needed to solve for angle (x). then solve for (x). 14. 15. 16. 17. 18.

Explanation:

Step1: Recall trigonometric ratios

For right - triangles, we use tangent ($\tan$), sine ($\sin$), or cosine ($\cos$) ratios. $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$.

Step2: Solve for problem 14

In right - triangle $ABC$, $\tan x=\frac{AC}{BC}=\frac{22}{15}$. Then $x = \arctan(\frac{22}{15})\approx55.8^{\circ}$.

Step3: Solve for problem 15

In right - triangle $DEF$, $\sin x=\frac{DE}{DF}=\frac{12}{68}=\frac{3}{17}$. Then $x=\arcsin(\frac{3}{17})\approx10.2^{\circ}$.

Step4: Solve for problem 16

In right - triangle $GHK$, $\tan x=\frac{HK}{GH}=\frac{53}{88}$. Then $x=\arctan(\frac{53}{88})\approx30.9^{\circ}$.

Step5: Solve for problem 17

In right - triangle $NML$, $\sin x=\frac{ML}{NL}=\frac{90}{230}=\frac{9}{23}$. Then $x=\arcsin(\frac{9}{23})\approx23.1^{\circ}$.

Step6: Solve for problem 18

In right - triangle $PQR$, $\tan x = 1$ since $\frac{QR}{PQ}=\frac{76}{76}=1$. Then $x = 45^{\circ}$.

Answer:

1. $x\approx55.8^{\circ}$
2. $x\approx10.2^{\circ}$
3. $x\approx30.9^{\circ}$
4. $x\approx23.1^{\circ}$
5. $x = 45^{\circ}$