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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 \(x=DE\) (opposite to \(\angle D\)). 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, \(\sin D=\frac{DE}{DF}\).

Step2: Substitute the values

Substitute \(D = 18^{\circ}\) and \(DF = 52\) into the sine formula. We get \(\sin(18^{\circ})=\frac{x}{52}\).

Step3: Solve for \(x\)

Multiply both sides of the equation by \(52\): \(x = 52\times\sin(18^{\circ})\).
We know that \(\sin(18^{\circ})\approx0.3090\). Then \(x = 52\times0.3090=16.068\).

Answer:

\(x\approx16.1\)