Question

1. triangle with right angle, 60° angle, hypotenuse labeled 3, legs labeled a (opposite 60°) and b (adjacent to 60°)

Explanation:

To solve for \(a\) and \(b\) in the right - angled triangle with a \(60^{\circ}\) angle and hypotenuse \(c = 3\), we can use trigonometric ratios.

Step 1: Analyze the angles of the triangle

In a right - angled triangle, the sum of the interior angles is \(180^{\circ}\). We know one angle is \(90^{\circ}\) and another is \(60^{\circ}\), so the third angle \(\theta=180^{\circ}-90^{\circ}-60^{\circ} = 30^{\circ}\).

Step 2: Use trigonometric ratios to find \(b\) (adjacent to \(60^{\circ}\) angle)

We know that \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\theta = 60^{\circ}\), \(\cos(60^{\circ})=\frac{b}{3}\). Since \(\cos(60^{\circ})=\frac{1}{2}\), we have the equation \(\frac{1}{2}=\frac{b}{3}\).
To solve for \(b\), we cross - multiply: \(b = 3\times\frac{1}{2}=\frac{3}{2}\)

Step 3: Use trigonometric ratios to find \(a\) (opposite to \(60^{\circ}\) angle)

We know that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\theta = 60^{\circ}\), \(\sin(60^{\circ})=\frac{a}{3}\). Since \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), we have the equation \(\frac{\sqrt{3}}{2}=\frac{a}{3}\).
To solve for \(a\), we cross - multiply: \(a = 3\times\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}\)

If we assume the problem is to find the lengths of \(a\) and \(b\):

• The length of \(b\) is \(\frac{3}{2}\)
• The length of \(a\) is \(\frac{3\sqrt{3}}{2}\)

If the problem is to find \(a\) or \(b\) (depending on the requirement), for example, if we want to find \(b\):

Answer:

\(\frac{3}{2}\)

If we want to find \(a\):