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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 the trigonometric ratio

In right - triangle \(LMN\) with right - angle at \(M\), we know the hypotenuse \(LN = 18\) and the side opposite to angle \(x\) is \(MN=9.8\). We use the sine ratio. The sine of an angle in a right - triangle is defined as \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). So, \(\sin x=\frac{MN}{LN}\).

Step2: Substitute the values

Substitute \(MN = 9.8\) and \(LN = 18\) into the sine formula. We get \(\sin x=\frac{9.8}{18}\).

Step3: Calculate the value of \(\sin x\)

\(\frac{9.8}{18}\approx0.5444\).

Step4: Find the angle \(x\)

To find \(x\), we use the inverse - sine function. \(x=\sin^{- 1}(0.5444)\). Using a calculator, \(x\approx33.0^{\circ}\).

Answer:

\(x\approx33.0^{\circ}\)