Question

lets do that a few more times and look for a pattern!
calculate the values of x & y in the following 30-60-90 triangles.

1. what are the values of x and y?

image of a 30-60-90 triangle with hypotenuse 10, right angle, 30° angle opposite side x, 60° angle opposite side y

1. what are the values of x & y?

image of a 30-60-90 triangle with hypotenuse 6, right angle, 30° angle opposite side x, 60° angle opposite side y

Explanation:

Step1: Recall 30-60-90 triangle ratios

In a 30-60-90 triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite \(30^\circ\) is the shortest (\(x\)), the side opposite \(60^\circ\) is \(x\sqrt{3}\), and the hypotenuse is \(2x\). Also, the side adjacent to \(30^\circ\) (longer leg) is \(x\sqrt{3}\), and adjacent to \(60^\circ\) (shorter leg) is \(x\).

Step2: Solve for problem 2

• Hypotenuse = 10. So, \(2x = 10 \implies x = 5\) (shorter leg, opposite \(30^\circ\)).
• Longer leg \(y = x\sqrt{3} = 5\sqrt{3}\) (or adjacent to \(30^\circ\), opposite \(60^\circ\)). Wait, no: Wait, in the triangle, the right angle, \(30^\circ\) at top, \(60^\circ\) at bottom. So:
• The side opposite \(30^\circ\) is \(x\) (bottom leg), side opposite \(60^\circ\) is \(y\) (left leg), hypotenuse 10.
• So, hypotenuse = \(2 \times\) shorter leg (opposite \(30^\circ\)). So shorter leg \(x = \frac{\text{hypotenuse}}{2} = \frac{10}{2} = 5\).
• Longer leg (opposite \(60^\circ\)) \(y = x\sqrt{3} = 5\sqrt{3}\)? Wait, no: Wait, in 30-60-90, sides: shorter leg (opposite 30°) = \(a\), longer leg (opposite 60°) = \(a\sqrt{3}\), hypotenuse = \(2a\). So in the triangle, angle at bottom is 60°, so the side opposite 60° is \(y\) (left leg), side opposite 30° is \(x\) (bottom leg). So:
• Hypotenuse = 10 = \(2a\) (where \(a\) is shorter leg, \(x\)). So \(a = 5\), so \(x = 5\), \(y = a\sqrt{3} = 5\sqrt{3}\)? Wait, no, wait: Wait, the left leg is adjacent to 30°, so it's the longer leg. Wait, maybe I mixed up. Let's re-express:
• Angles: right angle (90°), 30° (top), 60° (bottom). So:
• Side opposite 30°: bottom leg (\(x\)) → shorter leg.
• Side opposite 60°: left leg (\(y\)) → longer leg.
• Hypotenuse: 10.
• So, shorter leg \(x = \frac{\text{hypotenuse}}{2} = \frac{10}{2} = 5\).
• Longer leg \(y = x \times \sqrt{3} = 5\sqrt{3}\)? Wait, no, longer leg is \(a\sqrt{3}\) where \(a\) is shorter leg. So yes, \(y = 5\sqrt{3}\), \(x = 5\).

Step3: Solve for problem 3

• Hypotenuse = 6. So, shorter leg \(x = \frac{\text{hypotenuse}}{2} = \frac{6}{2} = 3\) (opposite 30°).
• Longer leg \(y = x\sqrt{3} = 3\sqrt{3}\) (opposite 60°).

Wait, but let's confirm:

For problem 2:

• Triangle with right angle, 30° at top, 60° at bottom. So:
• Side opposite 30°: \(x\) (bottom) → shorter leg.
• Side opposite 60°: \(y\) (left) → longer leg.
• Hypotenuse: 10.
• So, shorter leg \(x = \frac{10}{2} = 5\).
• Longer leg \(y = 5 \times \sqrt{3} = 5\sqrt{3}\).

For problem 3:

• Hypotenuse = 6.
• Shorter leg \(x = \frac{6}{2} = 3\).
• Longer leg \(y = 3 \times \sqrt{3} = 3\sqrt{3}\).

Answer:

Problem 2:

\(x = 5\), \(y = 5\sqrt{3}\)

Problem 3:

\(x = 3\), \(y = 3\sqrt{3}\)