Question

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

Explanation:

Step1: Find \( d \) in the 30-60-90 triangle

In a 30-60-90 right 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 the hypotenuse is twice the shortest side. Here, the hypotenuse is 8, and \( d \) is opposite the 60° angle? Wait, no—wait, the right angle, 30°, 60°: the side opposite 30° is half the hypotenuse. Wait, the hypotenuse is 8, so the side opposite 30° is 4, and the side opposite 60° (which is \( d \)) is \( 4\sqrt{3} \)? Wait, no, let's check the angles. The triangle has angles 30°, 60°, 90°, with hypotenuse 8. The side adjacent to 30° (wait, no, \( d \) is the base, adjacent to 60°? Wait, maybe better to use trigonometry. \( \sin(60^\circ) = \frac{d}{8} \)? Wait, no, \( \sin(60^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} \). Wait, the angle at the top is 30°, so the angle at the bottom left is 60°, so the side opposite 60° is \( d \), hypotenuse is 8. So \( \sin(60^\circ) = \frac{d}{8} \), so \( d = 8 \sin(60^\circ) = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} \). Wait, or maybe \( \cos(30^\circ) = \frac{d}{8} \), since \( \cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} \), and adjacent to 30° is \( d \). Yes, that's better. \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), so \( d = 8 \times \frac{\sqrt{3}}{2} = 4\sqrt{3} \).

Step2: Find \( h \) in the 45-45-90 triangle

In a 45-45-90 triangle, the legs are equal, and the hypotenuse is \( \text{leg} \times \sqrt{2} \). Here, the hypotenuse is 5, and \( h \) is a leg. So \( \sin(45^\circ) = \frac{h}{5} \), so \( h = 5 \sin(45^\circ) = 5 \times \frac{\sqrt{2}}{2} = \frac{5\sqrt{2}}{2} \)? Wait, no—wait, 45-45-90 triangle: legs are equal, hypotenuse \( = \text{leg} \times \sqrt{2} \), so \( \text{leg} = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \). Yes, that's correct.

Wait, but let's recheck the first triangle. The triangle with angles 30°, 60°, 90°: hypotenuse 8. The side opposite 30° is 4 (since 30° opposite side is half hypotenuse), then the other leg (d) is \( 4\sqrt{3} \), which matches the trigonometric calculation. So \( d = 4\sqrt{3} \).

For the second triangle, 45-45-90, hypotenuse 5, so each leg is \( \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \), so \( h = \frac{5\sqrt{2}}{2} \).

Wait, but maybe I mixed up the first triangle. Let's confirm: in a 30-60-90 triangle, sides are \( x \), \( x\sqrt{3} \), \( 2x \), where \( x \) is the shortest side (opposite 30°). So if hypotenuse is \( 2x = 8 \), then \( x = 4 \). Then the side opposite 60° is \( x\sqrt{3} = 4\sqrt{3} \), which is \( d \). So that's correct.

So \( d = 4\sqrt{3} \), \( h = \frac{5\sqrt{2}}{2} \).

Answer:

\( d = 4\sqrt{3} \)
\( h = \frac{5\sqrt{2}}{2} \)