Question

1. what are the values of x & y?

(right triangle with angle 30°, 60°, right angle, hypotenuse 18, vertical side y, horizontal side x)
5)
(right triangle with angle 30°, 60°, right angle, hypotenuse 14, vertical side y, horizontal side x)

Explanation:

Step1: Analyze Triangle Properties (30-60-90)

In a 30-60-90 right triangle, the sides are in the ratio \(1 : \sqrt{3} : 2\), where the side opposite \(30^\circ\) is the shortest (let's call it \(a\)), the side opposite \(60^\circ\) is \(a\sqrt{3}\), and the hypotenuse is \(2a\).

For Problem 4:

- Hypotenuse = 18. So, the side opposite \(30^\circ\) (which is \(x\) if we check the angles? Wait, no: Wait, the angle at the bottom is \(60^\circ\), so the side opposite \(30^\circ\) (the top angle) is \(x\)? Wait, no, let's label the triangle: Right angle at the bottom left, so the sides:
- Hypotenuse: 18 (opposite right angle)
- Angle at bottom right: \(60^\circ\), so the side adjacent to \(60^\circ\) is \(y\) (the vertical leg), and the side opposite \(60^\circ\) is \(x\) (the horizontal leg). Wait, no: In a right triangle, the legs are opposite the non-right angles. So:
- Angle at top: \(30^\circ\), so the side opposite \(30^\circ\) is \(x\) (horizontal leg, since it's opposite the \(30^\circ\) angle at the top).
- Angle at bottom right: \(60^\circ\), so the side opposite \(60^\circ\) is \(y\) (vertical leg)? Wait, no, the vertical leg is adjacent to the \(30^\circ\) angle. Wait, maybe better to recall: In 30-60-90, hypotenuse = \(2 \times\) shorter leg (opposite \(30^\circ\)).

So, hypotenuse = 18, so shorter leg (opposite \(30^\circ\)) is \(18 / 2 = 9\)? Wait, no, wait the angle at the top is \(30^\circ\), so the side opposite \(30^\circ\) is the horizontal leg \(x\)? Wait, no, the horizontal leg is opposite the \(30^\circ\) angle? Wait, the right angle is at the bottom left, so the vertices are: Top (A), bottom left (B, right angle), bottom right (C). So angle at A: \(30^\circ\), angle at C: \(60^\circ\), angle at B: \(90^\circ\). Then:
- Side opposite angle A (\(30^\circ\)): BC (horizontal leg, \(x\))
- Side opposite angle C (\(60^\circ\)): AB (vertical leg, \(y\))
- Hypotenuse: AC = 18.

So, in 30-60-90, hypotenuse = \(2 \times\) shorter leg (BC, since it's opposite \(30^\circ\)). So:
\(AC = 2 \times BC\) → \(18 = 2x\) → \(x = 9\). Then, the longer leg (AB, opposite \(60^\circ\)) is \(x\sqrt{3} = 9\sqrt{3}\)? Wait, no, wait: Wait, no, the side opposite \(60^\circ\) is \(y\)? Wait, no, angle at C is \(60^\circ\), so side opposite \(60^\circ\) is AB (vertical leg), which is \(y\). So in 30-60-90, longer leg (opposite \(60^\circ\)) is \(shorter leg \times \sqrt{3}\). So shorter leg is \(x = 9\) (opposite \(30^\circ\)), so longer leg \(y = 9\sqrt{3}\)? Wait, but hypotenuse is 18, so \(x = 9\), \(y = 9\sqrt{3}\)? Wait, no, wait: Wait, hypotenuse is 18, so shorter leg (opposite \(30^\circ\)) is \(18 / 2 = 9\) (so \(x = 9\)), then longer leg (opposite \(60^\circ\)) is \(9 \times \sqrt{3} = 9\sqrt{3}\) (so \(y = 9\sqrt{3}\))? Wait, but let's check with trigonometry:

Using cosine and sine:

- \(\cos(60^\circ) = \frac{adjacent}{hypotenuse} = \frac{y}{18}\). Since \(\cos(60^\circ) = 0.5\), so \(y = 18 \times 0.5 = 9\)? Wait, that's conflicting. Wait, no, I messed up the angle labels.

Wait, angle at C is \(60^\circ\), so the adjacent side to angle C is \(y\) (vertical leg), and the opposite side is \(x\) (horizontal leg). So:

- \(\cos(60^\circ) = \frac{adjacent}{hypotenuse} = \frac{y}{18}\) → \(y = 18 \times \cos(60^\circ) = 18 \times 0.5 = 9\)
- \(\sin(60^\circ) = \frac{opposite}{hypotenuse} = \frac{x}{18}\) → \(x = 18 \times \sin(60^\circ) = 18 \times \frac{\sqrt{3}}{2} = 9\sqrt{3}\)

Ah, I see, I had the legs reversed. So the vertical leg \(y\) is adjacent to the \(60^\circ\) angle, so \(y = 9\), and the horizontal leg \(x = 9\sqrt{3}\). Wait, but that contradicts the 30-60-90 ratio. Wait, no: The angle at the top is \(30^\circ\), so the side opposite \(30^\circ\) is \(x\) (horizontal leg), because angle at top is \(30^\circ\), so opposite side is \(x\). Then hypotenuse is \(2x\), so \(2x = 18\) → \(x = 9\). Then the other leg (opposite \(60^\circ\)) is \(x\sqrt{3} = 9\sqrt{3}\), which is \(y\)? Wait, now I'm confused. Let's clarify:

In a right triangle, the side opposite \(30^\circ\) is the shortest leg. So if the top angle is \(30^\circ\), then the side opposite to it (BC, horizontal leg) is the shortest leg. So hypotenuse (AC) is twice that: \(AC = 2 \times BC\) → \(18 = 2x\) → \(x = 9\). Then the other leg (AB, vertical leg) is \(x\sqrt{3} = 9\sqrt{3}\), which is \(y\). But when we use cosine for angle C (\(60^\circ\)): \(\cos(60^\circ) = \frac{AB}{AC}\) (adjacent over hypotenuse) → \(0.5 = \frac{y}{18}\) → \(y = 9\). Wait, that's a contradiction. So where is the mistake?

Ah! Wait, the right angle is at B (bottom left), so the sides:

- AB: vertical leg (length \(y\))
- BC: horizontal leg (length \(x\))
- AC: hypotenuse (length 18)

Angle at A: \(30^\circ\), so angle at A is \(30^\circ\), so side opposite angle A is BC (\(x\)), so \(x\) is opposite \(30^\circ\), so \(x = \frac{AC}{2} = \frac{18}{2} = 9\). Then angle at C is \(60^\circ\), so side opposite angle C is AB (\(y\)), so \(y = x\sqrt{3} = 9\sqrt{3}\). But when we use cosine for angle C: \(\cos(60^\circ) = \frac{BC}{AC}\)? No, \(\cos(angle) = \frac{adjacent}{hypotenuse}\). Angle at C is \(60^\circ\), so adjacent side to angle C is BC (\(x\)), and opposite side is AB (\(y\)). So \(\cos(60^\circ) = \frac{BC}{AC} = \frac{x}{18}\) → \(0.5 = \frac{9}{18}\) → \(0.5 = 0.5\), which works. And \(\sin(60^\circ) = \frac{AB}{AC} = \frac{y}{18}\) → \(\frac{\sqrt{3}}{2} = \frac{y}{18}\) → \(y = 9\sqrt{3}\). So that's correct. My earlier mistake was in identifying adjacent and opposite sides for angle C. So angle C is \(60^\circ\), adjacent side is BC (\(x\)), opposite is AB (\(y\)). So that's consistent.

So for problem 4:

- \(x = 9\) (opposite \(30^\circ\), so shorter leg)
- \(y = 9\sqrt{3}\) (opposite \(60^\circ\), longer leg)

For Problem 5:

- Hypotenuse = 14. So, using the 30-60-90 ratio:

Shorter leg (opposite \(30^\circ\)) is \(x\) (if angle at top is \(30^\circ\), so side opposite is \(x\)), so hypotenuse = \(2x\) → \(14 = 2x\) → \(x = 7\). Then the longer leg (opposite \(60^\circ\)) is \(x\sqrt{3} = 7\sqrt{3}\), which is \(y\).

Wait, let's confirm with trigonometry:

Angle at top: \(30^\circ\), so side opposite \(30^\circ\) is \(x\) (horizontal leg), so \(x = \frac{14}{2} = 7\). Then vertical leg \(y\) (opposite \(60^\circ\)) is \(7\sqrt{3}\).

Step2: Summarize the Results

For problem 4:
- \(x = 9\)
- \(y = 9\sqrt{3}\) (or approximately \(15.588\))

For problem 5:
- \(x = 7\)
- \(y = 7\sqrt{3}\) (or approximately \(12.124\))

Answer:

Problem 4:

\(x = 9\), \(y = 9\sqrt{3}\) (or \(x = 9\), \(y \approx 15.59\))

Problem 5:

\(x = 7\), \(y = 7\sqrt{3}\) (or \(x = 7\), \(y \approx 12.12\))