Question

find x and y

Explanation:

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

• Since one angle is \(45^{\circ}\) and it is a right - triangle, it is a \(45 - 45-90\) triangle. In a \(45 - 45 - 90\) triangle, the legs are equal. So \(x = 3\).
• The hypotenuse \(y\) in a \(45 - 45 - 90\) triangle is related to the leg \(a\) by the formula \(y=a\sqrt{2}\). Substituting \(a = 3\), we get \(y = 3\sqrt{2}\).

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

• We know that \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\) and \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Given \(\theta = 46^{\circ}\) and the opposite side \(BC = 4\).
• Using \(\tan46^{\circ}=\frac{4}{x}\), we can solve for \(x\). \(x=\frac{4}{\tan46^{\circ}}\approx\frac{4}{1.0355}\approx3.86\).
• Using \(\cos46^{\circ}=\frac{x}{y}\), and \(x\approx3.86\), we have \(y=\frac{x}{\cos46^{\circ}}\). Since \(\cos46^{\circ}\approx0.6947\), \(y=\frac{3.86}{0.6947}\approx5.56\).

Answer:

For the first triangle: \(x = 3\), \(y = 3\sqrt{2}\approx4.24\). For the second triangle: \(x\approx3.86\), \(y\approx5.56\).