Question

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

Explanation:

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

In a 45-45-90 right triangle, the legs are equal, and the hypotenuse \( c = a\sqrt{2} \) (where \( a \) is a leg). Here, hypotenuse is 6, and let leg \( b \) be \( x \). So \( 6 = x\sqrt{2} \), solving for \( x \): \( x=\frac{6}{\sqrt{2}}=\frac{6\sqrt{2}}{2}=3\sqrt{2} \)? Wait, no—wait, in a 45-45-90 triangle, legs are equal. Wait, the angles are 45, 45, 90, so it's an isosceles right triangle. So the two legs (one is \( b \), the other is the vertical leg) are equal. The hypotenuse is 6. So by Pythagoras: \( b^2 + b^2 = 6^2 \) → \( 2b^2 = 36 \) → \( b^2 = 18 \) → \( b = \sqrt{18} = 3\sqrt{2} \)? Wait, no, wait—wait, maybe I mixed up. Wait, the hypotenuse is opposite the right angle. Wait, the triangle has angles 45, 45, 90, so the legs are equal. Let me re-express: Let the legs be \( b \) (horizontal) and \( y \) (vertical), so \( b = y \). Hypotenuse \( c = 6 \). Then \( b^2 + y^2 = c^2 \) → \( 2b^2 = 36 \) → \( b^2 = 18 \) → \( b = 3\sqrt{2} \). Wait, but maybe the hypotenuse is 6, and the legs are \( b \) and the other leg. Wait, no—wait, the triangle is drawn with the horizontal leg \( b \), vertical leg (let's say \( h \)), hypotenuse 6, angles 45 and 45. So yes, \( b = h \), so \( b = \frac{6}{\sqrt{2}} = 3\sqrt{2} \). Wait, but maybe I made a mistake. Wait, no—alternatively, in 45-45-90, leg = hypotenuse / \( \sqrt{2} \). So \( b = \frac{6}{\sqrt{2}} = 3\sqrt{2} \).

Wait, but maybe the triangle is such that the hypotenuse is 6, and the legs are \( b \) and the other leg. So yes, \( b = 3\sqrt{2} \).

Step2: Find \( d \) (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 (let's say \( x \)), opposite 60° is \( x\sqrt{3} \), and hypotenuse is \( 2x \). Here, the angle 60° is opposite side \( d \), and the side opposite 30° is 7? Wait, no—wait, the right angle, angle 60°, angle 30°. So the side opposite 30° is the shortest side. Let's identify the angles: right angle (90°), 60°, 30°. So the side opposite 30° is the side with length 7? Wait, the horizontal side is 7 (adjacent to 60°), vertical side is \( d \) (opposite 60°), hypotenuse is the slant side. Wait, in a 30-60-90 triangle, the side opposite 30° is the shortest, let's call it \( x \), side opposite 60° is \( x\sqrt{3} \), hypotenuse \( 2x \). So here, the angle 30° is at the top, 60° at the bottom left, right angle at bottom right. So the side opposite 30° is the horizontal side (length 7), because angle 30° is at the top, so the side opposite is the bottom side (length 7). So \( x = 7 \) (opposite 30°), then the side opposite 60° (which is \( d \)) is \( x\sqrt{3} = 7\sqrt{3} \)? Wait, no—wait, angle at the bottom left is 60°, so the side opposite 60° is \( d \) (vertical side), and the side opposite 30° is the horizontal side (length 7). So in 30-60-90 triangle:
- Opposite 30°: \( x \) (horizontal, length 7)
- Opposite 60°: \( x\sqrt{3} \) (vertical, \( d \))
- Hypotenuse: \( 2x \)

So \( d = x\sqrt{3} = 7\sqrt{3} \)? Wait, no—wait, tangent of 60° is \( \frac{d}{7} \), since \( \tan(60°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{d}{7} \). \( \tan(60°) = \sqrt{3} \), so \( d = 7\sqrt{3} \). Alternatively, using 30-60-90 ratios: if the side opposite 30° is \( x \), then side opposite 60° is \( x\sqrt{3} \), hypotenuse \( 2x \). Here, side opposite 30° is 7 (since angle 30° is at the top, so opposite side is the bottom side, length 7). So \( x = 7 \), so side opposite 60° ( \( d \)) is \( 7\sqrt{3} \).

Wait, let's recheck:

For the first triangle (45-45-90):
- Angles: 45, 45, 90. So it's isosceles right triangle. Hypotenuse = 6. Legs are equal. So leg \( b = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2} \).

For the second triangle (30-60-90):
- Angles: 30, 60, 90. The side adjacent to 60° is 7 (horizontal), side opposite 60° is \( d \) (vertical). So \( \tan(60°) = \frac{d}{7} \) → \( d = 7\tan(60°) = 7\sqrt{3} \). Alternatively, using cosine: \( \cos(60°) = \frac{7}{\text{hypotenuse}} \), but we need \( d \), so sine: \( \sin(60°) = \frac{d}{\text{hypotenuse}} \), but maybe tangent is easier. Since \( \tan(60°) = \sqrt{3} \), so \( d = 7\sqrt{3} \).

Wait, but let's confirm 30-60-90 ratios. In a 30-60-90 triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \), where the side opposite 30° is the shortest (1 part), opposite 60° is \( \sqrt{3} \) parts, hypotenuse 2 parts. So if the side opposite 30° is 7 (the horizontal side, since angle 30° is at the top, so opposite side is the bottom side, length 7), then the side opposite 60° (vertical side \( d \)) is \( 7\sqrt{3} \), and hypotenuse is \( 14 \). That makes sense.

So:

- For \( b \) (45-45-90 triangle): hypotenuse 6, legs equal. So \( b = \frac{6}{\sqrt{2}} = 3\sqrt{2} \). Wait, but wait—maybe I had the hypotenuse wrong. Wait, in the first triangle, the hypotenuse is 6, and the legs are \( b \) and the other leg. So by Pythagoras, \( b^2 + b^2 = 6^2 \) → \( 2b^2 = 36 \) → \( b^2 = 18 \) → \( b = \sqrt{18} = 3\sqrt{2} \). Correct.

- For \( d \) (30-60-90 triangle): side opposite 60° is \( d \), side opposite 30° is 7. So \( d = 7\sqrt{3} \) (since \( \tan(60°) = \sqrt{3} = \frac{d}{7} \) → \( d = 7\sqrt{3} \)).

Wait, but let's check the first triangle again. Maybe the hypotenuse is not 6? Wait, the first triangle has angles 45, 45, 90, with hypotenuse 6. So legs are equal. So yes, \( b = 3\sqrt{2} \).

Wait, but maybe I made a mistake in the 45-45-90 triangle. Let's recall: in a 45-45-90 triangle, leg length \( = \frac{\text{hypotenuse}}{\sqrt{2}} \). So hypotenuse 6, leg \( = \frac{6}{\sqrt{2}} = 3\sqrt{2} \). Correct.

For the 30-60-90 triangle: angle 60° at the bottom, so adjacent side is 7 (horizontal), opposite side is \( d \) (vertical). So \( \tan(60°) = \frac{d}{7} \) → \( d = 7\tan(60°) = 7\sqrt{3} \). Correct.

So:

\( b = 3\sqrt{2} \)? Wait, no—wait, wait a second. Wait, in the first triangle, maybe the hypotenuse is not 6? Wait, the first triangle is drawn with the hypotenuse as 6, and the two legs: one is \( b \) (horizontal), the other is vertical (same as \( b \) because 45-45-90). So yes, \( b = 3\sqrt{2} \).

Wait, but let's re-express:

First triangle (45-45-90):

- Angles: 45°, 45°, 90°
- Hypotenuse: 6
- Legs: \( b \) (horizontal) and \( h \) (vertical), \( b = h \)
- By Pythagoras: \( b^2 + h^2 = 6^2 \) → \( 2b^2 = 36 \) → \( b^2 = 18 \) → \( b = \sqrt{18} = 3\sqrt{2} \)

Second triangle (30-60-90):

- Angles: 30°, 60°, 90°
- Side adjacent to 60°: 7 (horizontal)
- Side opposite to 60°: \( d \) (vertical)
- \( \tan(60°) = \frac{\text{opposite}}{\text{adjacent}} = \frac{d}{7} \)
- \( \tan(60°) = \sqrt{3} \) → \( d = 7\sqrt{3} \)

Yes, that seems correct.

Answer:

\( b = 3\sqrt{2} \), \( d = 7\sqrt{3} \)