Question

for the right triangles below, find the exact values of the side lengths c and b. if necessary, write your responses in simplified radical form. (there are two right triangles: one with angles 45°, 45°, 90° and leg length 2, hypotenuse c; another with angles 30°, 60°, 90°, one leg length 6, hypotenuse b)

Explanation:

Step1: Find \( c \) (45-45-90 triangle)

In a 45-45-90 right triangle, the legs are equal, and the hypotenuse \( c \) is \( \text{leg} \times \sqrt{2} \). The leg is 2, so \( c = 2\sqrt{2} \)? Wait, no—wait, the leg is 2, so hypotenuse \( c = 2\sqrt{2} \)? Wait, no, 45-45-90 triangle: legs are equal, hypotenuse \( = \text{leg} \times \sqrt{2} \). Wait, the leg here is 2, so \( c = 2\sqrt{2} \)? Wait, no, wait the first triangle (left) has angles 45°, 45°, 90°, and one leg is 2. So hypotenuse \( c = 2\sqrt{2} \)? Wait, no, wait 45-45-90 triangle: if leg is \( a \), hypotenuse is \( a\sqrt{2} \). So if leg is 2, hypotenuse \( c = 2\sqrt{2} \)? Wait, no, maybe I misread. Wait the left triangle: right angle, one angle 45°, so it's isoceles right triangle. So both legs are equal. The horizontal leg is 2, so the vertical leg is also 2. Then hypotenuse \( c = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \). Correct.

Step2: Find \( b \) (30-60-90 triangle)

In a 30-60-90 right triangle, the sides are in ratio \( 1 : \sqrt{3} : 2 \), where the side opposite 30° is the shortest (let's call it \( x \)), opposite 60° is \( x\sqrt{3} \), and hypotenuse is \( 2x \). Wait, in the right triangle (right), the angle 30° is adjacent to the side of length 6? Wait, no: the right triangle has angles 30°, 60°, 90°. The side of length 6: let's see, the angle 30°—wait, the side opposite 30° would be the shortest. Wait, the side labeled 6: is it opposite 60° or 30°? Let's check: the angle at the top is 30°, so the side opposite 30° is the horizontal leg? Wait, no, the right angle is at the bottom, so the vertical leg is 6, the angle at the top is 30°, so the side opposite 30° is the horizontal leg (let's call it \( x \)), the vertical leg (opposite 60°) is \( x\sqrt{3} = 6 \), so \( x = \frac{6}{\sqrt{3}} = 2\sqrt{3} \), and hypotenuse \( b = 2x = 4\sqrt{3} \)? Wait, no, wait: in 30-60-90 triangle, the sides are: opposite 30°: \( x \), opposite 60°: \( x\sqrt{3} \), hypotenuse: \( 2x \). So if the vertical leg (opposite 60°) is 6, then \( x\sqrt{3} = 6 \implies x = \frac{6}{\sqrt{3}} = 2\sqrt{3} \). Then hypotenuse \( b = 2x = 4\sqrt{3} \)? Wait, no, wait the hypotenuse is \( b \), and the angle at the top is 30°, so the side opposite 30° is the horizontal leg (length \( x \)), the vertical leg (opposite 60°) is \( x\sqrt{3} = 6 \), so \( x = 2\sqrt{3} \), hypotenuse \( b = 2x = 4\sqrt{3} \)? Wait, but maybe I got the angles wrong. Wait the triangle has angles 30°, 60°, 90°, right angle at bottom. So the angle at the top is 30°, so the side opposite 30° is the horizontal leg (adjacent to 60°), and the vertical leg (opposite 60°) is 6. So yes, \( x\sqrt{3} = 6 \implies x = 2\sqrt{3} \), hypotenuse \( b = 2x = 4\sqrt{3} \). Wait, but let's recheck: if hypotenuse is \( b \), angle 30° at top, so \( \cos(30°) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{6}{b} \implies b = \frac{6}{\cos(30°)} = \frac{6}{\frac{\sqrt{3}}{2}} = 6 \times \frac{2}{\sqrt{3}} = \frac{12}{\sqrt{3}} = 4\sqrt{3} \). Correct. And \( \sin(30°) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\text{horizontal leg}}{b} \implies \text{horizontal leg} = b \times \sin(30°) = 4\sqrt{3} \times \frac{1}{2} = 2\sqrt{3} \), which matches the earlier calculation.

Wait, but the first triangle (left) is 45-45-90, leg 2, so hypotenuse \( c = 2\sqrt{2} \)? Wait, no, wait the left triangle: the horizontal leg is 2, right angle, 45° angle, so it's isoceles right triangle, so both legs are 2, hypotenuse \( c = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \). Correct.

Wait, but the problem says "for the right triangles below, find the exact values of the side lengths \( c \) and \( b \)". So \( c \) is for the left triangle (45-45-90), \( b \) is for the right triangle (30-60-90).

So:

For \( c \) (left triangle, 45-45-90):

Leg length = 2, so hypotenuse \( c = 2\sqrt{2} \)? Wait, no, wait 45-45-90 triangle: hypotenuse = leg × √2. So leg is 2, so \( c = 2\sqrt{2} \).

For \( b \) (right triangle, 30-60-90):

The side opposite 60° is 6, so hypotenuse \( b = \frac{6}{\sin(60°)} \). Since \( \sin(60°) = \frac{\sqrt{3}}{2} \), so \( b = \frac{6}{\frac{\sqrt{3}}{2}} = 6 \times \frac{2}{\sqrt{3}} = 4\sqrt{3} \). Alternatively, using the ratio: in 30-60-90, hypotenuse is twice the side opposite 30°. The side opposite 30° is \( \frac{6}{\sqrt{3}} = 2\sqrt{3} \), so hypotenuse \( b = 2 \times 2\sqrt{3} = 4\sqrt{3} \).

Wait, but let's confirm:

Left triangle (45-45-90):

- Angles: 45°, 45°, 90°
- Legs: both 2 (since one leg is 2, and it's isoceles)
- Hypotenuse \( c = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \)

Right triangle (30-60-90):

- Angles: 30°, 60°, 90°
- Side opposite 60°: 6 (vertical leg)
- In 30-60-90, side opposite 60° is \( x\sqrt{3} \), where \( x \) is side opposite 30°
- So \( x\sqrt{3} = 6 \implies x = \frac{6}{\sqrt{3}} = 2\sqrt{3} \)
- Hypotenuse \( b = 2x = 2 \times 2\sqrt{3} = 4\sqrt{3} \)

Yes, that makes sense.

Wait, but maybe I made a mistake with \( c \). Wait the left triangle: the horizontal leg is 2, right angle, 45° angle, so the other leg (vertical) is also 2, so hypotenuse \( c = \sqrt{2^2 + 2^2} = \sqrt{8} = 2\sqrt{2} \). Correct.

So:

\( c = 2\sqrt{2} \)

\( b = 4\sqrt{3} \)

Wait, but let's check again. Wait the left triangle: the angle is 45°, right angle, so it's isoceles, so both legs are equal. The horizontal leg is 2, so vertical leg is 2, hypotenuse \( c = 2\sqrt{2} \). Correct.

Right triangle: angle at top is 30°, right angle at bottom, so the vertical leg is opposite 60°, so length 6. So hypotenuse \( b = \frac{6}{\cos(30°)} \)? Wait, no, cos(30°) is adjacent over hypotenuse. The adjacent to 30° is the vertical leg (6), so \( \cos(30°) = \frac{6}{b} \implies b = \frac{6}{\cos(30°)} = \frac{6}{\frac{\sqrt{3}}{2}} = 4\sqrt{3} \). Correct.

So final answers:

\( c = 2\sqrt{2} \)

\( b = 4\sqrt{3} \)

Wait, but let's check if the left triangle's leg is 2. The left triangle has a horizontal leg of 2, right angle, 45° angle, so yes, it's isoceles right triangle, so hypotenuse \( c = 2\sqrt{2} \).

Yes.

Answer:

\( c = 2\sqrt{2} \)

\( b = 4\sqrt{3} \)