Question

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

Explanation:

Step1: Find \( h \) in the 45-45-90 triangle

In a 45-45-90 right triangle, the legs are equal, and the hypotenuse \( c = \text{leg} \times \sqrt{2} \). Here, the hypotenuse is 3, and the leg \( h \) can be found using \( \sin(45^\circ) = \frac{h}{3} \). Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), we have \( h = 3 \times \frac{\sqrt{2}}{2} \)? Wait, no—wait, in a 45-45-90 triangle, the legs are equal, and hypotenuse \( = \text{leg} \times \sqrt{2} \). So if hypotenuse is 3, then leg \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \)? Wait, no, wait the triangle has angles 45-45-90, so it's isoceles right triangle. So the two legs are equal, and hypotenuse is leg \( \times \sqrt{2} \). So if hypotenuse is 3, then leg \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \)? Wait, no, maybe I misread. Wait the first triangle: hypotenuse is 3, angles 45-45-90, so the legs are equal. So \( \sin(45^\circ) = \frac{h}{3} \), so \( h = 3 \sin(45^\circ) = 3 \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \). Wait, but maybe it's a 45-45-90 triangle with leg 3? Wait no, the side labeled 3 is the hypotenuse? Wait the first triangle: the side labeled 3 is the hypotenuse, and the angle is 45 degrees, so it's a 45-45-90 triangle. So legs are equal, hypotenuse \( = \text{leg} \times \sqrt{2} \), so leg \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \). Wait, but maybe I made a mistake. Wait the second triangle: angles 30-60-90, with the shorter leg (opposite 30 degrees) being 4? Wait no, in a 30-60-90 triangle, the sides are in ratio \( 1 : \sqrt{3} : 2 \), where the side opposite 30° is the shortest leg (length \( x \)), opposite 60° is \( x\sqrt{3} \), hypotenuse is \( 2x \). So the second triangle: the right angle, 60° at the bottom, 30° at the top. So the side adjacent to 60° is 4? Wait no, the side labeled 4 is adjacent to 60°? Wait the second triangle: right angle, 60° at the bottom, 30° at the top, so the side opposite 30° is the shorter leg. Wait the side labeled 4: is it the side opposite 30° or 60°? Wait, angle at the bottom is 60°, so the side opposite 60° is \( a \), and the side opposite 30° is 4? Wait no, in a 30-60-90 triangle, the side opposite 30° is the shortest leg (length \( x \)), hypotenuse is \( 2x \), and side opposite 60° is \( x\sqrt{3} \). So if the angle at the bottom is 60°, then the side adjacent to 60° is the side opposite 30°, which is 4? Wait, no. Let's label the second triangle: right angle, angle at bottom is 60°, angle at top is 30°, so the side opposite 30° is the side adjacent to 60°, which is 4? Wait, no. Let's denote: in triangle with angles 30°, 60°, 90°, the side opposite 30° is the shortest leg (let's call it \( x \)), side opposite 60° is \( x\sqrt{3} \), hypotenuse is \( 2x \). So in the second triangle, the side labeled 4 is adjacent to the 60° angle, so it's the side opposite 30°, so \( x = 4 \)? Wait no, adjacent to 60° is the side opposite 30°, so if angle at bottom is 60°, then the side adjacent to 60° is the vertical leg? Wait no, the right angle is at the bottom right, so the horizontal leg is 4, angle at bottom left is 60°, angle at top is 30°. So the horizontal leg (length 4) is adjacent to the 60° angle, so it's the side opposite the 30° angle. So in 30-60-90 triangle, side opposite 30° is \( x \), side opposite 60° is \( x\sqrt{3} \), hypotenuse \( 2x \). So if side opposite 30° is 4, then hypotenuse \( h \) (wait no, the hypotenuse is the side opposite the right angle, so the hypotenuse of the second triangle is the side opposite the right angle, which is the longest side. Wait the second triangle: right angle at bottom right, angle at bottom left 60°, angle at top 30°, horizontal leg (bottom) is 4, vertical leg is \( a \), hypotenuse is the side from top to bottom left. So the side opposite 30° is the horizontal leg (length 4), because 30° is at the top, so opposite side is the horizontal leg (length 4). So in 30-60-90 triangle, side opposite 30° is \( x = 4 \), so hypotenuse \( = 2x = 8 \)? Wait no, hypotenuse is opposite the right angle, so hypotenuse is the side from top vertex to bottom left vertex. So side opposite 30° (top angle) is the horizontal leg (length 4), so \( x = 4 \), so hypotenuse \( = 2x = 8 \)? Wait no, that can't be. Wait, no—wait, in a 30-60-90 triangle, the side opposite 30° is the shortest side, so if the horizontal leg is 4, and it's opposite 30°, then hypotenuse is \( 2 \times 4 = 8 \), and the vertical leg (opposite 60°) is \( 4\sqrt{3} \). Wait, but the vertical leg is \( a \), so \( a = 4\sqrt{3} \), and hypotenuse (the side from top to bottom left) is 8? Wait, no, maybe I mixed up. Wait let's re-express:

First triangle (45-45-90):

- Angles: 45°, 45°, 90°
- Hypotenuse: 3 (the side labeled 3)
- Legs: equal, let's call each leg \( h \)
- In 45-45-90 triangle, hypotenuse \( = h\sqrt{2} \), so \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \)

Second triangle (30-60-90):

- Angles: 30°, 60°, 90°
- Horizontal leg (length 4) is opposite the 30° angle (since 30° is at the top, so opposite side is horizontal leg)
- In 30-60-90 triangle, side opposite 30° is \( x \), side opposite 60° is \( x\sqrt{3} \), hypotenuse \( 2x \)
- So \( x = 4 \) (side opposite 30°), so hypotenuse (the side from top to bottom left) is \( 2x = 8 \)? Wait no, hypotenuse is the side opposite the right angle, so it's the side connecting the top vertex to the bottom left vertex. The vertical leg (length \( a \)) is opposite the 60° angle, so \( a = x\sqrt{3} = 4\sqrt{3} \), and hypotenuse \( = 2x = 8 \). Wait, but the problem says "find the exact values of the side lengths \( h \) and \( a \)". So \( h \) is from the first triangle, \( a \) from the second.

Wait, let's confirm:

First triangle: 45-45-90, hypotenuse 3. So legs are equal. So \( \cos(45^\circ) = \frac{h}{3} \), so \( h = 3 \cos(45^\circ) = 3 \times \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \).

Second triangle: 30-60-90, horizontal leg 4 (opposite 30°), so vertical leg \( a \) (opposite 60°) is \( 4\sqrt{3} \), and hypotenuse (but the problem asks for \( a \), which is the vertical leg). Wait, no—wait the second triangle: the vertical leg is \( a \), horizontal leg is 4, angle at bottom left 60°, so the side opposite 60° is \( a \), and side opposite 30° is 4. So in 30-60-90, side opposite 30° is \( x \), side opposite 60° is \( x\sqrt{3} \), so \( x = 4 \), so \( a = 4\sqrt{3} \), and hypotenuse is \( 2x = 8 \). But the problem asks for \( h \) (from first triangle) and \( a \) (from second triangle).

Wait, maybe I messed up the first triangle. Wait the first triangle: the side labeled 3 is a leg, not hypotenuse? Wait the first triangle: angles 45-45-90, so it's isoceles right triangle, so the two legs are equal, and hypotenuse is leg \( \times \sqrt{2} \). If the side labeled 3 is a leg, then hypotenuse is \( 3\sqrt{2} \), but the problem says "find \( h \)". Wait the first triangle: the side labeled 3 is the hypotenuse? Or a leg? Let's look at the diagram: first triangle, hypotenuse is 3, legs are \( h \) and the other leg (equal to \( h \)) because 45-45-90. So \( h \) is a leg, hypotenuse 3. So \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \).

Second triangle: 30-60-90, horizontal leg 4 (adjacent to 60° angle, so opposite 30° angle), so vertical leg \( a \) is opposite 60° angle, so \( a = 4\sqrt{3} \), and hypotenuse is 8. But the problem asks for \( h \) (from first triangle) and \( a \) (from second triangle).

Wait, let's re-express:

First triangle (45-45-90):

- Hypotenuse: 3
- Legs: \( h \) (equal, since 45-45-90)
- So \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \) (rationalized)

Second triangle (30-60-90):

- Horizontal leg: 4 (opposite 30° angle, so shortest side)
- Vertical leg: \( a \) (opposite 60° angle, so longer leg)
- In 30-60-90, longer leg = shorter leg \( \times \sqrt{3} \)
- So \( a = 4 \times \sqrt{3} = 4\sqrt{3} \)

Wait, but maybe the first triangle's side labeled 3 is a leg, not hypotenuse. Let's check: if 3 is a leg, then hypotenuse is \( 3\sqrt{2} \), but the problem says "find \( h \)". If \( h \) is a leg, and 3 is a leg, then \( h = 3 \). But that would be if it's a 45-45-90 with leg 3. But the diagram shows the side labeled 3 as the hypotenuse? Wait the user's diagram: first triangle, hypotenuse is 3, legs are \( h \) and the other leg (equal to \( h \)) because 45-45-90. So \( h \) is a leg, hypotenuse 3. So \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \).

Second triangle: 30-60-90, horizontal leg 4 (adjacent to 60° angle, so opposite 30° angle), so vertical leg \( a \) is opposite 60° angle, so \( a = 4\sqrt{3} \).

So putting it together:

\( h = \frac{3\sqrt{2}}{2} \)? Wait no, wait maybe I made a mistake. Wait the first triangle: angles 45-45-90, so it's an isoceles right triangle, so the two legs are equal, and hypotenuse is leg \( \times \sqrt{2} \). So if the hypotenuse is 3, then leg \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \).

Second triangle: 30-60-90, the side opposite 30° is 4 (horizontal leg), so the side opposite 60° (vertical leg \( a \)) is \( 4\sqrt{3} \), and hypotenuse is 8.

So the answers are \( h = \frac{3\sqrt{2}}{2} \) and \( a = 4\sqrt{3} \)? Wait no, wait the second triangle: if the horizontal leg is 4, and it's the side adjacent to the 30° angle, then maybe I got the angles wrong. Wait the second triangle: angle at top is 30°, angle at bottom left is 60°, right angle at bottom right. So the side opposite 30° is the horizontal leg (length 4), side opposite 60° is the vertical leg (length \( a \)), hypotenuse is the side from top to bottom left. So in 30-60-90, side opposite 30° is \( x \), side opposite 60° is \( x\sqrt{3} \), hypotenuse \( 2x \). So \( x = 4 \), so \( a = 4\sqrt{3} \), hypotenuse \( 8 \).

First triangle: 45-45-90, hypotenuse 3, so leg \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \).

Yes, that makes sense.

Step1: Solve for \( h \) (45-45-90 triangle)

In a 45-45-90 right triangle, the legs are equal, and hypotenuse \( = \text{leg} \times \sqrt{2} \). Let the leg length be \( h \), hypotenuse \( = 3 \).
Thus, \( 3 = h \times \sqrt{2} \), so \( h = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \) (rationalizing the denominator).

Step2: Solve for \( a \) (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), where the shorter leg is opposite the 30° angle.
Here, the shorter leg (opposite 30°) is 4 (horizontal leg). The longer leg (opposite 60°) is \( a \), so \( a = \text{shorter leg} \times \sqrt{3} = 4\sqrt{3} \).

Answer:

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