Question

solve for ( x ). round to the nearest tenth, if necessary.
(there is a right triangle ( fgh ) with right angle at ( g ), ( fh = 47 ), ( angle h = 26^circ ), and ( gh = x ))
answer attempt 1 out of 2
( x = ) input box

Explanation:

Step1: Identify trigonometric ratio

We use tangent: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$
Here, $\theta=26^\circ$, opposite side to $\angle H$ is $x$, adjacent side is 47.

Step2: Rearrange to solve for $x$

$x = 47 \times \tan(26^\circ)$

Step3: Calculate the value

First find $\tan(26^\circ)\approx0.4877$, then $x\approx47\times0.4877$
$x\approx22.9219$

Step4: Round to nearest tenth

$x\approx22.9$

Answer:

$x=22.9$