Question

unit 4: ratios, sides, & angles ... cash belcher solve for the side set up an equation, then solve for x. be accurate to within one decimal.

Explanation:

Step1: Use sine function

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. Here, $\theta = 15^{\circ}$, the hypotenuse is $11$ and the side we want to find $x$ is the opposite side to the angle $15^{\circ}$. So, $\sin(15^{\circ})=\frac{x}{11}$.

Step2: Solve for $x$

We know that $\sin(15^{\circ})=\sin(45^{\circ}- 30^{\circ})=\sin45^{\circ}\cos30^{\circ}-\cos45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}\approx0.259$. Then $x = 11\times\sin(15^{\circ})$. Substituting the value of $\sin(15^{\circ})$, we get $x=11\times0.259 = 2.849\approx2.9$.

Answer:

$x \approx 2.9$