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Question
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10
60°
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Step1: Use sine function for \(x\)
In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here \(\theta = 60^{\circ}\), hypotenuse \(= 10\), and the side opposite to \(60^{\circ}\) is \(x\).
\(\sin60^{\circ}=\frac{x}{10}\)
Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), we have \(\frac{\sqrt{3}}{2}=\frac{x}{10}\), then \(x = 10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\)
Step2: Use cosine function for \(y\)
In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here \(\theta = 60^{\circ}\), hypotenuse \(= 10\), and the side adjacent to \(60^{\circ}\) is \(y\).
\(\cos60^{\circ}=\frac{y}{10}\)
Since \(\cos60^{\circ}=\frac{1}{2}\), we have \(\frac{1}{2}=\frac{y}{10}\), then \(y = 10\times\frac{1}{2}=5\)
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\(x = 5\sqrt{3}\), \(y = 5\)