Question

solve for ( x ). round to the nearest tenth, if necessary.
(image of a right triangle ( fgh ) with right angle at ( g ), ( fh = x ), ( gh = 23 ), ( angle f = 71^circ ))
answer attempt 1 out of 2

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( FGH \), \( \angle F = 71^\circ \), hypotenuse \( FH = x \), and opposite side to \( \angle F \) is \( HG = 23 \). Use sine: \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \). So \( \sin(71^\circ)=\frac{23}{x} \).

Step2: Solve for \( x \)

Rearrange: \( x = \frac{23}{\sin(71^\circ)} \). Calculate \( \sin(71^\circ)\approx0.9455 \). Then \( x\approx\frac{23}{0.9455}\approx24.3 \).

Answer:

\( 24.3 \)