Question

for the right triangles below, find the exact values of the side lengths ( d ) and ( b ). if necessary, write your responses in simplified radical form.

Explanation:

Step1: Solve for \( d \) (30-60-90 triangle)

In a 30-60-90 triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \), where the side opposite 30° is the shortest, opposite 60° is \( \sqrt{3} \) times that, and hypotenuse is twice the shortest. Here, hypotenuse is 5, so the side opposite 30° (let's say \( x \)) is \( \frac{5}{2} \), and \( d \) (opposite 60°) is \( x\sqrt{3} \). So \( d = \frac{5}{2}\sqrt{3}=\frac{5\sqrt{3}}{2} \). Wait, no—wait, the right angle, 60°, 30°: the side adjacent to 30° is \( d \)? Wait, no, let's label: right angle, 60° at top, 30° at bottom, hypotenuse 5. So the side opposite 30° is \( d \)? Wait, no, 30° angle: the side opposite 30° is the shorter leg. Wait, maybe I mixed up. Let's use trigonometry. \( \sin(30^\circ)=\frac{d}{5} \)? No, \( \sin(60^\circ)=\frac{d}{5} \)? Wait, angle at bottom is 30°, so the side opposite 30° is the vertical leg (d)? Wait, no, right triangle: angles 90°, 60°, 30°. So the side opposite 30° is the shorter leg, opposite 60° is longer leg, hypotenuse. So if hypotenuse is 5, then shorter leg (opposite 30°) is \( \frac{5}{2} \), longer leg (opposite 60°) is \( \frac{5\sqrt{3}}{2} \). Wait, but \( d \) is adjacent to 30°? Wait, maybe better to use cosine: \( \cos(30^\circ)=\frac{d}{5} \)? No, \( \cos(30^\circ)=\frac{\text{adjacent}}{\text{hypotenuse}} \). If 30° angle, adjacent side is \( d \)? Wait, no, let's draw: right angle at left, 60° at top, 30° at bottom. So the sides: left leg is \( d \) (vertical), bottom leg is adjacent to 30°, hypotenuse 5. So \( \sin(30^\circ)=\frac{d}{5} \)? No, \( \sin(30^\circ)=\frac{\text{opposite}}{\text{hypotenuse}} \). Opposite to 30° is \( d \)? Wait, no, 30° is at bottom, so opposite side is \( d \) (left leg). So \( \sin(30^\circ)=\frac{d}{5} \), so \( d = 5 \times \sin(30^\circ) = 5 \times \frac{1}{2} = \frac{5}{2} \)? Wait, that contradicts earlier. Wait, no, 60° at top: opposite side is bottom leg. Wait, I'm confused. Let's use the ratio: 30-60-90 triangle: sides are \( x \), \( x\sqrt{3} \), \( 2x \), where \( x \) is opposite 30°, \( x\sqrt{3} \) opposite 60°, \( 2x \) hypotenuse. So if hypotenuse is 5, then \( 2x = 5 \implies x = \frac{5}{2} \). So the side opposite 30° is \( x = \frac{5}{2} \), opposite 60° is \( x\sqrt{3} = \frac{5\sqrt{3}}{2} \). Now, which angle is which? The triangle has 60° at top, 30° at bottom, right angle at left. So the side opposite 30° (bottom angle) is the left leg (d)? Wait, no, bottom angle is 30°, so opposite side is left leg (d). Then \( d = x = \frac{5}{2} \)? Wait, no, that can't be. Wait, maybe the hypotenuse is 5, and the angle at bottom is 30°, so the side adjacent to 30° is the bottom leg, and opposite is \( d \). So \( \sin(30^\circ) = \frac{d}{5} \implies d = 5 \times \frac{1}{2} = \frac{5}{2} \). Then the other leg (bottom) is \( \sqrt{5^2 - (\frac{5}{2})^2} = \sqrt{\frac{100}{4} - \frac{25}{4}} = \sqrt{\frac{75}{4}} = \frac{5\sqrt{3}}{2} \). Okay, so \( d = \frac{5}{2} \)? Wait, no, maybe I had the angle wrong. Let's check the second triangle: 45-45-90, which is isoceles, so legs are equal. So for the 45-45-90 triangle, hypotenuse is 8, so legs \( b \) and the other leg are equal. In 45-45-90, legs are \( x \), \( x \), hypotenuse \( x\sqrt{2} \). So \( x\sqrt{2} = 8 \implies x = \frac{8}{\sqrt{2}} = 4\sqrt{2} \). So \( b = 4\sqrt{2} \).

Step2: Solve for \( b \) (45-45-90 triangle)

In a 45-45-90 triangle, the legs are equal, and hypotenuse \( = \text{leg} \times \sqrt{2} \). Given hypotenuse is 8, let leg be \( b \). So \( b\sqrt{2} = 8 \implies b = \frac{8}{\sqrt{2}} = 4\sqrt{2} \) (rationalizing the denominator: \( \frac{8\sqrt{2}}{2} = 4\sqrt{2} \)).

Wait, earlier mistake with \( d \): let's re-examine the 30-60-90 triangle. Angles: 90°, 60°, 30°. So the side opposite 30° is the shorter leg, opposite 60° is longer leg, hypotenuse. So if hypotenuse is 5, shorter leg (opposite 30°) is \( \frac{5}{2} \), longer leg (opposite 60°) is \( \frac{5\sqrt{3}}{2} \). Now, in the diagram, the 30° angle is at the bottom, 60° at the top, right angle at left. So the side \( d \) is adjacent to 30°? No, \( d \) is the vertical leg (left), which is opposite 60°? Wait, 60° is at top, so opposite side is the bottom leg. Wait, I'm getting confused. Let's use trigonometry: \( \cos(60^\circ) = \frac{d}{5} \). \( \cos(60^\circ) = 0.5 \), so \( d = 5 \times 0.5 = 2.5 = \frac{5}{2} \). Yes, that's correct. Because \( \cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} \), and \( d \) is adjacent to 60° (since 60° is at top, adjacent side is \( d \), hypotenuse 5). So \( d = 5 \times \cos(60^\circ) = 5 \times \frac{1}{2} = \frac{5}{2} \).

Answer:

\( d = \frac{5}{2} \)
\( b = 4\sqrt{2} \)