Question

solve for ( x ). round to the nearest tenth of a degree, if necessary.
triangle hij with right angle at i, hi = 9, hj = 18, angle at j is ( x^circ )

Explanation:

Step1: Identify the trigonometric ratio

In a right - triangle, the sine of an angle is defined as $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. For angle $x$, the opposite side is $HI = 9$ and the hypotenuse is $HJ=18$.
So, $\sin x=\frac{HI}{HJ}$.

Step2: Substitute the values

Substitute $HI = 9$ and $HJ = 18$ into the formula: $\sin x=\frac{9}{18}=\frac{1}{2}$.

Step3: Solve for $x$

We know that if $\sin x = a$, then $x=\sin^{- 1}(a)$. Since $\sin x=\frac{1}{2}$, then $x=\sin^{-1}(\frac{1}{2})$.
Using the inverse - sine function, $x = 30^{\circ}$ (because $\sin30^{\circ}=\frac{1}{2}$).

Answer:

$30^{\circ}$