Question

for the right triangles below, find the exact values of the side lengths a and c. if necessary, write your responses in simplified radical form. (images of two right triangles: first with angles 30°, 60°, 90° and one leg 2, side a; second with angles 45°, 45°, 90° and one leg 4, hypotenuse c) a = c =

Explanation:

Step1: Find \( a \) (30-60-90 triangle)

In a 30-60-90 right triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \) (shorter leg : longer leg : hypotenuse). The longer leg (opposite 60°) is \( \sqrt{3} \) times the shorter leg, and the hypotenuse is twice the shorter leg. Here, the longer leg (adjacent to 30°) is 2? Wait, no—wait, the angle of 30°: the side opposite 30° is the shorter leg. Wait, the right angle, 60°, 30°: so the side opposite 30° is \( a \)? Wait no, the side labeled 2 is adjacent to 30°? Wait, no, let's re-examine. The first triangle: right angle, 60°, 30°. So the sides: the side opposite 30° is the shorter leg, opposite 60° is longer leg, hypotenuse. Wait, the side labeled 2: is that the longer leg (opposite 60°) or the shorter leg? Wait, in 30-60-90, tan(30°) = opposite/adjacent = \( a / 2 \)? Wait, no, angle 30°: adjacent side is 2, opposite is \( a \), right angle. So tan(30°) = \( a / 2 \). Tan(30°) is \( \frac{1}{\sqrt{3}} \), so \( a = 2 \times \frac{1}{\sqrt{3}} \)? No, wait, maybe I got the angles wrong. Wait, the triangle has angles 90°, 60°, 30°, so the side opposite 30° is the shortest side. Wait, maybe the side labeled 2 is the longer leg (opposite 60°), so the shorter leg (opposite 30°) is \( a \), and longer leg is \( a\sqrt{3} \). So if longer leg is 2, then \( a\sqrt{3} = 2 \), so \( a = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \)? No, that doesn't seem right. Wait, no, maybe the side labeled 2 is the hypotenuse? No, hypotenuse would be the longest side. Wait, maybe I mixed up. Alternatively, cot(30°) = adjacent/opposite = 2 / a. Cot(30°) is \( \sqrt{3} \), so \( \sqrt{3} = 2 / a \), so \( a = 2 / \sqrt{3} = \frac{2\sqrt{3}}{3} \)? Wait, no, maybe the first triangle: angle 30°, right angle, so the side opposite 30° is \( a \), and the side adjacent (the leg with length 2) is the longer leg (opposite 60°). So in 30-60-90, longer leg = shorter leg × \( \sqrt{3} \). So if longer leg is 2, then shorter leg (a) = \( 2 / \sqrt{3} = \frac{2\sqrt{3}}{3} \)? Wait, but maybe I made a mistake. Wait, let's check the second triangle: 45-45-90, so it's an isosceles right triangle, so the legs are equal, and hypotenuse is leg × \( \sqrt{2} \). The leg is 4, so hypotenuse \( c = 4\sqrt{2} \). That's straightforward. So for the second triangle: 45-45-90, legs are 4 and 4 (since it's isosceles), so hypotenuse \( c = 4\sqrt{2} \). Now back to first triangle: 30-60-90. Let's use trigonometry. Angle 30°, adjacent side is 2, opposite is \( a \), right angle. So tan(30°) = \( a / 2 \). Tan(30°) = \( 1/\sqrt{3} \), so \( a = 2 \times (1/\sqrt{3}) = 2/\sqrt{3} = 2\sqrt{3}/3 \)? Wait, but maybe the side labeled 2 is the hypotenuse? No, hypotenuse would be longer than both legs. Wait, maybe the first triangle: the side opposite 30° is \( a \), and the side adjacent (the other leg) is 2, and hypotenuse is \( 2a \) (since in 30-60-90, hypotenuse is twice the shorter leg). Then, by Pythagoras: \( a^2 + 2^2 = (2a)^2 \). So \( a^2 + 4 = 4a^2 \) → \( 3a^2 = 4 \) → \( a^2 = 4/3 \) → \( a = 2/\sqrt{3} = 2\sqrt{3}/3 \). Yes, that matches. So \( a = 2\sqrt{3}/3 \).

Step2: 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 4, so \( c = 4\sqrt{2} \).

Answer:

\( a = \frac{2\sqrt{3}}{3} \), \( c = 4\sqrt{2} \)