Question

question 5: standard g.srt.6
use the sine ratio to solve for the missing side (x).
right triangle with hypotenuse 10, angle 30°, leg x
a. 7
b. 5
c. 10
d. 4

question 6: standard g.srt.6
use the sine ratio to solve for the missing side (x).
right triangle with leg 50, angle 45°, leg x
a. 35.5
b. 50
c. 70.7
d. 12.4

question 7: standard g.srt.6
use the cosine ratio to solve for the missing side (x).
right triangle with hypotenuse 12, angle 40°, leg x
a. 9.2
b. 13.6
c. 10
d. 21.5

Explanation:

Question 5

Step1: Recall sine ratio

In a right triangle, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = 30^\circ$, opposite side is $x$, hypotenuse is $10$. So $\sin(30^\circ) = \frac{x}{10}$.

Step2: Solve for x

We know $\sin(30^\circ) = 0.5$, so $0.5 = \frac{x}{10}$. Multiply both sides by 10: $x = 10 \times 0.5 = 5$.

Step1: Recall sine ratio

$\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = 45^\circ$, opposite side is $50$, hypotenuse is $x$. So $\sin(45^\circ) = \frac{50}{x}$.

Step2: Solve for x

$\sin(45^\circ) = \frac{\sqrt{2}}{2} \approx 0.707$. So $0.707 = \frac{50}{x}$. Then $x = \frac{50}{0.707} \approx 70.7$.

Step1: Recall cosine ratio

$\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$. Here, $\theta = 40^\circ$, adjacent side is $x$, hypotenuse is $12$. So $\cos(40^\circ) = \frac{x}{12}$.

Step2: Solve for x

$\cos(40^\circ) \approx 0.766$. So $x = 12 \times 0.766 \approx 9.2$.

Answer:

b. 5

Question 6