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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 trigonometric ratio

In right - triangle $QRS$ with right - angle at $R$, we use the sine ratio. $\sin(x)=\frac{\text{opposite}}{\text{hypotenuse}}$. Here, the side opposite to angle $x$ is $3.8$ and the hypotenuse is $6.7$. So, $\sin(x)=\frac{3.8}{6.7}$.

Step2: Solve for $x$

We take the inverse - sine of both sides: $x = \sin^{-1}(\frac{3.8}{6.7})$.
Using a calculator, $\frac{3.8}{6.7}\approx0.5672$. Then $x=\sin^{-1}(0.5672)$.
$x\approx34.6^{\circ}$ (rounded to the nearest tenth of a degree).

Answer:

$34.6$