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QUESTION IMAGE

solve for x. round to the nearest tenth of a degree, if necessary.

Question

solve for x. round to the nearest tenth of a degree, if necessary.

Explanation:

Step1: Identify the trigonometric ratio

In the right triangle \( \triangle HIJ \) (right - angled at \( I \)), we know the length of the opposite side to angle \( x \) (which is \( HI = 9 \)) and the length of the hypotenuse (which is \( HJ=18 \)). The sine of an angle in a right triangle is defined as \( \sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}} \). So, for angle \( x \), we have \( \sin(x)=\frac{HI}{HJ} \).

Substituting the given values: \( \sin(x)=\frac{9}{18}=\frac{1}{2} \)

Step2: Solve for \( x \)

We know that if \( \sin(x)=\frac{1}{2} \), and we are looking for an angle \( x \) in a triangle (so \( 0^{\circ}

Answer:

\( 30.0^{\circ} \) (rounded to the nearest tenth of a degree)