Question

applying the 30°-60°-90° triangle theorem
in a triangle whose angle measures are 30°-60°-90°, one leg of the triangle is 80 cm, as shown in the diagram. how long are the remaining two sides of the triangle?
(diagram of triangle with right angle at g, angle 60° at h, side gj = 80, and some equations: 1:√3:2, (√3)/2 = x/?, (√3 x)/√3 = (160√3)/√3, x = (160√3)/3)

Explanation:

Step1: Identify Triangle Type

The triangle is \(30^\circ\)-\(60^\circ\)-\(90^\circ\), so side ratios are \(1:\sqrt{3}:2\) (shorter leg : longer leg : hypotenuse).

Step2: Analyze Given Side

Side \(GJ = 80\). Let's check which leg it is. Angle at \(H\) is \(60^\circ\), so \(\angle J = 30^\circ\). Thus, \(GJ\) (opposite \(60^\circ\)) is the longer leg, shorter leg \(GH\), hypotenuse \(JH\).

Step3: Find Shorter Leg (\(GH\))

Let shorter leg \(= x\), longer leg \(= x\sqrt{3}\). Given longer leg \(= 80\), so \(x\sqrt{3}=80 \Rightarrow x = \frac{80}{\sqrt{3}}=\frac{80\sqrt{3}}{3}\)? Wait, no—wait the diagram's calculation: Wait, maybe \(GJ = 80\) is adjacent to \(30^\circ\)? Wait, re-express: In \(30^\circ\)-\(60^\circ\)-\(90^\circ\), if \(\angle J = 30^\circ\), then hypotenuse \(JH\), shorter leg \(GH\) (opposite \(30^\circ\)), longer leg \(GJ\) (opposite \(60^\circ\)). So longer leg \(= \text{shorter leg} \times \sqrt{3}\), hypotenuse \(= 2 \times \text{shorter leg}\).
Wait, the given calculation has \(\frac{\sqrt{3}}{2}=\frac{80}{x}\)? Wait, maybe the hypotenuse is \(x\)? Wait, let's correct: Let hypotenuse \(JH = x\), longer leg \(GJ = 80\). In ratio, \(\frac{\text{longer leg}}{\text{hypotenuse}}=\frac{\sqrt{3}}{2}\), so \(\frac{80}{x}=\frac{\sqrt{3}}{2} \Rightarrow x = \frac{160}{\sqrt{3}}=\frac{160\sqrt{3}}{3}\)? No, wait the calculation in the diagram: \(\frac{\sqrt{3}}{2}=\frac{80}{x}\)? Wait, no—wait the diagram's step: \(\frac{\sqrt{3}}{2}=\frac{80}{x}\)? Wait, solving \(\frac{\sqrt{3}}{2}=\frac{80}{x}\) gives \(x = \frac{160}{\sqrt{3}}=\frac{160\sqrt{3}}{3}\)? No, wait the diagram has \(x = \frac{160\sqrt{3}}{3}\)? Wait, no—wait the correct ratio: shorter leg \(= s\), longer leg \(= s\sqrt{3}\), hypotenuse \(= 2s\). If longer leg \(= 80\), then \(s\sqrt{3}=80 \Rightarrow s = \frac{80}{\sqrt{3}}=\frac{80\sqrt{3}}{3}\) (shorter leg \(GH\)), hypotenuse \(JH = 2s = \frac{160\sqrt{3}}{3}\)? No, that can't be. Wait, maybe the given side is the shorter leg? Wait, the diagram's calculation shows \(x = \frac{160\sqrt{3}}{3}\), which is hypotenuse? Wait, no—let's re-express the problem.

Wait, the problem says "one leg is 80 cm". Let's assume: In \(30^\circ\)-\(60^\circ\)-\(90^\circ\), legs are \(s\) (shorter, opposite \(30^\circ\)) and \(s\sqrt{3}\) (longer, opposite \(60^\circ\)), hypotenuse \(2s\).

Case 1: 80 is the shorter leg (\(s = 80\)):
- Longer leg \(= 80\sqrt{3}\)
- Hypotenuse \(= 160\)

Case 2: 80 is the longer leg (\(s\sqrt{3}=80 \Rightarrow s = \frac{80}{\sqrt{3}}=\frac{80\sqrt{3}}{3}\)):
- Shorter leg \(= \frac{80\sqrt{3}}{3}\)
- Hypotenuse \(= \frac{160\sqrt{3}}{3}\)

But the diagram's calculation has \(x = \frac{160\sqrt{3}}{3}\), which would be hypotenuse if 80 is longer leg. Wait, the diagram's step: \(\frac{\sqrt{3}}{2}=\frac{80}{x}\) (longer leg / hypotenuse \(= \frac{\sqrt{3}}{2}\)), so \(x = \frac{160}{\sqrt{3}}=\frac{160\sqrt{3}}{3}\) (hypotenuse), and shorter leg \(= \frac{80}{\sqrt{3}}=\frac{80\sqrt{3}}{3}\)? Wait, no—shorter leg is \(s\), hypotenuse \(2s\), longer leg \(s\sqrt{3}\). So if longer leg \(= 80 = s\sqrt{3} \Rightarrow s = \frac{80}{\sqrt{3}}\), hypotenuse \(= 2s = \frac{160}{\sqrt{3}}=\frac{160\sqrt{3}}{3}\), which matches the diagram's \(x = \frac{160\sqrt{3}}{3}\) (hypotenuse), and shorter leg \(= \frac{80}{\sqrt{3}}=\frac{80\sqrt{3}}{3}\).

Wait, but the problem says "one leg is 80 cm". So:
- If 80 is the longer leg (opposite \(60^\circ\)):
- Shorter leg: \(\frac{80\sqrt{3}}{3}\) cm
- Hypotenuse: \(\frac{160\sqrt{3}}{3}\) cm

- If 80 is the shorter leg (opposite \(30^\circ\)):
- Longer leg: \(80\sqrt{3}\) cm
- Hypotenuse: \(160\) cm

But the diagram's calculation uses \(\frac{\sqrt{3}}{2}=\frac{80}{x}\), implying longer leg / hypotenuse \(= \frac{\sqrt{3}}{2}\), so 80 is longer leg, hypotenuse \(x = \frac{160\sqrt{3}}{3}\), shorter leg \(= \frac{80\sqrt{3}}{3}\).

Wait, maybe the diagram has a typo, but following the calculation:

From \(\frac{\sqrt{3}}{2}=\frac{80}{x}\), cross-multiplying: \(\sqrt{3}x = 160 \Rightarrow x = \frac{160}{\sqrt{3}}=\frac{160\sqrt{3}}{3}\) (hypotenuse), and shorter leg: since longer leg \(= 80 = \text{shorter leg} \times \sqrt{3} \Rightarrow \text{shorter leg} = \frac{80}{\sqrt{3}}=\frac{80\sqrt{3}}{3}\).

Answer:

If the 80 cm leg is the longer leg (opposite \(60^\circ\)):

• Shorter leg: \(\boldsymbol{\frac{80\sqrt{3}}{3}}\) cm
• Hypotenuse: \(\boldsymbol{\frac{160\sqrt{3}}{3}}\) cm

If the 80 cm leg is the shorter leg (opposite \(30^\circ\)):

• Longer leg: \(\boldsymbol{80\sqrt{3}}\) cm
• Hypotenuse: \(\boldsymbol{160}\) cm

(But based on the diagram’s calculation, the hypotenuse is \(\frac{160\sqrt{3}}{3}\) and shorter leg \(\frac{80\sqrt{3}}{3}\), assuming 80 is the longer leg.)