Question

solve for ( x ). round to the nearest tenth, if necessary.
(there is a right triangle ( \triangle nom ) with right angle at ( n ), ( no = 7 ), ( angle o = 52^circ ), and ( nm = x ))
answer
( x = ) blank for answer

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( \triangle ONM \), \( \angle O = 52^\circ \), adjacent side to \( \angle O \) is \( ON = 7 \), and opposite side is \( MN = x \). We use the tangent function: \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \).
So, \( \tan(52^\circ)=\frac{x}{7} \).

Step2: Solve for \( x \)

Multiply both sides by 7: \( x = 7\times\tan(52^\circ) \).
Calculate \( \tan(52^\circ)\approx1.2799 \), then \( x = 7\times1.2799\approx8.9593 \).
Round to nearest tenth: \( x\approx9.0 \).

Answer:

\( x \approx 9.0 \)