Question

solve for ( x ). round to the nearest tenth, if necessary.
right triangle ( opq ) with right angle at ( p ), ( op = x ), ( pq = 17 ), and ( angle poq = 56^circ )

Explanation:

Step1: Identify trigonometric ratio

We have a right triangle at $P$. For $\angle O = 56^\circ$, side $x$ is adjacent, and side $PQ=17$ is opposite. We use the tangent ratio: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$
$\tan(56^\circ)=\frac{17}{x}$

Step2: Rearrange to solve for $x$

Isolate $x$ by cross-multiplying:
$x=\frac{17}{\tan(56^\circ)}$

Step3: Calculate the value

Use $\tan(56^\circ)\approx1.4826$:
$x\approx\frac{17}{1.4826}\approx11.5$

Answer:

$11.5$