Question

unit exam - right triangles and trigonometry
find the value of x.
$9\sqrt{3}$
$x = \boldsymbol{?}$

Explanation:

Step1: Identify trigonometric ratio

We use cosine: $\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$, where $\theta=60^\circ$, adjacent side $=9\sqrt{3}$, hypotenuse $=x$.

Step2: Substitute values into formula

$\cos(60^\circ)=\frac{9\sqrt{3}}{x}$
We know $\cos(60^\circ)=\frac{1}{2}$, so:
$\frac{1}{2}=\frac{9\sqrt{3}}{x}$

Step3: Solve for x

Cross-multiply to isolate $x$:
$x=2\times9\sqrt{3}$

Answer:

$18\sqrt{3}$