Question

question 8: standard g.srt.6
geometry
use the cosine ratio to solve for the missing side (x).
right triangle with hypotenuse 15, angle 25° at the base, base side x
a. 10
b. 3.9
c. 13.6
d. 4.3
question 9: standard g.srt.6
use the cosine ratio to solve for the missing side (x).
right triangle with hypotenuse 20, angle 60° at the top, vertical side x
a. 9.2
b. 13.6
c. 10
d. 21.5
question 10: standard g.srt.7
sin 40 degrees = cos ____ degrees
a. 20
b. 30
c. 40
d. 50
question 11: standard g.srt.7
cos 45 degrees = sin ____ degrees
a. 35
b. 25
c. 45
d. 55

Explanation:

Question 8

Step1: Recall cosine ratio

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$. Here, $\theta = 25^\circ$, hypotenuse = 15, adjacent = $x$.
So, $\cos(25^\circ) = \frac{x}{15}$

Step2: Solve for $x$

$x = 15 \times \cos(25^\circ)$
$\cos(25^\circ) \approx 0.9063$
$x \approx 15 \times 0.9063 \approx 13.6$

Step1: Recall cosine ratio

In a right triangle, $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}}$. Here, $\theta = 60^\circ$, hypotenuse = 20, adjacent = $x$.
So, $\cos(60^\circ) = \frac{x}{20}$

Step2: Solve for $x$

$\cos(60^\circ) = 0.5$
$x = 20 \times 0.5 = 10$

We know the co - function identity $\sin\alpha=\cos(90^{\circ}-\alpha)$.
Let $\sin(40^{\circ})=\cos(x)$. Then $x = 90^{\circ}- 40^{\circ}=50^{\circ}$

Answer:

c. 13.6

Question 9