Question

which equation could be used to find m∠j in △jkl? x = sin⁻¹(8.8/11) x = sin⁻¹(11/8.8) x = cos⁻¹(8.8/11) x = cos⁻¹(11/8.8)

Explanation:

Step1: Recall sine - cosine relations in right - triangle.

In right - triangle $\triangle{JKL}$, $\sin x=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos x=\frac{\text{adjacent}}{\text{hypotenuse}}$. Here, $\sin x=\frac{KL}{JK}=\frac{8.8}{11}$, so $x = \sin^{-1}(\frac{8.8}{11})$.

Answer:

$x=\sin^{-1}(\frac{8.8}{11})$