Question

solve for ( x ). round to the nearest tenth of a degree, if necessary.
right triangle mlk with right angle at l, hypotenuse mk = 92, leg lk = 69, angle at m is ( x^circ )

Explanation:

Step1: Identify the trigonometric ratio

In right - triangle \(MLK\) with \(\angle L = 90^{\circ}\), we know the opposite side (\(LK = 69\)) and the hypotenuse (\(MK=92\)). We use the sine ratio. The sine of an angle \(x\) in a right - triangle is defined as \(\sin x=\frac{\text{opposite}}{\text{hypotenuse}}\). So, \(\sin x=\frac{LK}{MK}\).

Step2: Substitute the values

Substitute \(LK = 69\) and \(MK = 92\) into the sine formula. We get \(\sin x=\frac{69}{92}\).

Step3: Simplify the fraction

\(\frac{69}{92}=\frac{3}{4}=0.75\). So, \(\sin x = 0.75\).

Step4: Find the angle \(x\)

To find \(x\), we use the inverse - sine function. \(x=\sin^{- 1}(0.75)\). Using a calculator, \(x\approx48.6^{\circ}\).

Answer:

\(48.6^{\circ}\)