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QUESTION IMAGE

solve for x. round to the nearest tenth, if necessary.

Question

solve for x. round to the nearest tenth, if necessary.

Explanation:

Step1: Identify the trigonometric ratio

In right - triangle \(DEF\) with \(\angle D = 18^{\circ}\), hypotenuse \(DF = 52\), and we want to find the side \(x=EF\) (opposite to \(\angle D\)). Use the sine ratio: \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\).
So, \(\sin(18^{\circ})=\frac{x}{52}\).

Step2: Solve for \(x\)

Multiply both sides of the equation \(\sin(18^{\circ})=\frac{x}{52}\) by \(52\).
\(x = 52\times\sin(18^{\circ})\).
Since \(\sin(18^{\circ})\approx0.3090\), then \(x = 52\times0.3090\).
\(x=16.068\).

Answer:

\(x\approx16.1\)