Question

directions
to complete this assignment you will be answering a series of questions centered ar
choose the response that answers the question best.
feeling stuck on a question?

• remember the ratios for solving these triangles. simplify the ratios as much as possi

applicable.

• revisit video lessons in 5.2 for a review of these topics.

question 1
solve for x and y.
16
60°
$x = 16,y = 8\sqrt{2}$
$x = 16,y = 8\sqrt{3}$
$x = 8,y = 8\sqrt{3}$
$x = 8,y = 8\sqrt{3}$

Explanation:

Step1: Use cosine function for x

In a right - triangle, $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Here, $\theta = 60^{\circ}$ and hypotenuse is 16. So, $\cos60^{\circ}=\frac{x}{16}$. Since $\cos60^{\circ}=\frac{1}{2}$, we have $x = 16\times\cos60^{\circ}=16\times\frac{1}{2}=8$.

Step2: Use sine function for y

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = 60^{\circ}$ and hypotenuse is 16. So, $\sin60^{\circ}=\frac{y}{16}$. Since $\sin60^{\circ}=\frac{\sqrt{3}}{2}$, we have $y = 16\times\sin60^{\circ}=16\times\frac{\sqrt{3}}{2}=8\sqrt{3}$.

Answer:

$x = 8,y = 8\sqrt{3}$