# Explanation: ## Step1: Solve for \(x\) and \(y\) in the first triangle - For the first right - triangle with hypotenuse \(AB = 10\) and angle \(A=60^{\circ}\): - Use the trigonometric ratios. \(\sin60^{\circ}=\frac{y}{10}\), so \(y = 10\times\sin60^{\circ}=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\). - \(\cos60^{\circ}=\frac{x}{10}\), so \(x = 10\times\cos60^{\circ}=10\times\frac{1}{2}=5\). ## Step2: Solve for \(x\) and \(y\) in the second triangle - For the second right - triangle with hypotenuse \(AB = 20\) and angle \(A = 30^{\circ}\): - Use the trigonometric ratios. \(\sin30^{\circ}=\frac{x}{20}\), so \(x=20\times\sin30^{\circ}=20\times\frac{1}{2} = 10\). - \(\cos30^{\circ}=\frac{y}{20}\), so \(y = 20\times\cos30^{\circ}=20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\). # Answer: For the first triangle: \(x = 5,y = 5\sqrt{3}\); for the second triangle: \(x = 10,y = 10\sqrt{3}\)

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QUESTION IMAGE

Question

find x and y

Explanation:

Step1: Solve for \(x\) and \(y\) in the first triangle

For the first right - triangle with hypotenuse \(AB = 10\) and angle \(A=60^{\circ}\):
Use the trigonometric ratios. \(\sin60^{\circ}=\frac{y}{10}\), so \(y = 10\times\sin60^{\circ}=10\times\frac{\sqrt{3}}{2}=5\sqrt{3}\).
\(\cos60^{\circ}=\frac{x}{10}\), so \(x = 10\times\cos60^{\circ}=10\times\frac{1}{2}=5\).

Step2: Solve for \(x\) and \(y\) in the second triangle

For the second right - triangle with hypotenuse \(AB = 20\) and angle \(A = 30^{\circ}\):
Use the trigonometric ratios. \(\sin30^{\circ}=\frac{x}{20}\), so \(x=20\times\sin30^{\circ}=20\times\frac{1}{2} = 10\).
\(\cos30^{\circ}=\frac{y}{20}\), so \(y = 20\times\cos30^{\circ}=20\times\frac{\sqrt{3}}{2}=10\sqrt{3}\).

Answer:

For the first triangle: \(x = 5,y = 5\sqrt{3}\); for the second triangle: \(x = 10,y = 10\sqrt{3}\)