Question

13)
15)
solve each triangle. round
17)
19)
21)
23)

Answer:

Let's solve each triangle one by one (we'll take problem 13 as an example first, and you can follow the same method for others):

Problem 13:

We have a right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \), \( BC = 4 \), \( \angle A = 41^\circ \), and we need to find \( x = AC \).

Step 1: Identify the trigonometric ratio

In a right triangle, \( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \). Here, for \( \angle A \), the opposite side is \( BC = 4 \), and the adjacent side is \( AC = x \). So:
\[
\tan(41^\circ) = \frac{BC}{AC} = \frac{4}{x}
\]

Step 2: Solve for \( x \)

\[
x = \frac{4}{\tan(41^\circ)}
\]
Using a calculator, \( \tan(41^\circ) \approx 0.8693 \):
\[
x \approx \frac{4}{0.8693} \approx 4.6
\]

Problem 15:

Right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \), hypotenuse \( AB = 10.3 \), \( \angle B = 37^\circ \), find \( x = AC \).

Step 1: Trigonometric ratio

\( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \). For \( \angle B \), opposite side is \( AC = x \), hypotenuse \( AB = 10.3 \):
\[
\sin(37^\circ) = \frac{AC}{AB} = \frac{x}{10.3}
\]

Step 2: Solve for \( x \)

\[
x = 10.3 \times \sin(37^\circ)
\]
\( \sin(37^\circ) \approx 0.6018 \):
\[
x \approx 10.3 \times 0.6018 \approx 6.2
\]

Problem 17:

Right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \), \( AC = 22.6 \) mi, \( \angle B = 62^\circ \).

Step 1: Find \( \angle A \)

In a right triangle, \( \angle A + \angle B = 90^\circ \):
\[
\angle A = 90^\circ - 62^\circ = 28^\circ
\]

Step 2: Find \( BC \) (let's call it \( y \))

\( \tan(62^\circ) = \frac{AC}{BC} \):
\[
\tan(62^\circ) = \frac{22.6}{y} \implies y = \frac{22.6}{\tan(62^\circ)}
\]
\( \tan(62^\circ) \approx 1.8807 \):
\[
y \approx \frac{22.6}{1.8807} \approx 12.0 \text{ mi}
\]

Step 3: Find \( AB \) (hypotenuse, let's call it \( z \))

\( \cos(62^\circ) = \frac{BC}{AB} \implies AB = \frac{BC}{\cos(62^\circ)} \), or \( \sin(62^\circ) = \frac{AC}{AB} \implies AB = \frac{22.6}{\sin(62^\circ)} \)
\( \sin(62^\circ) \approx 0.8829 \):
\[
AB \approx \frac{22.6}{0.8829} \approx 25.6 \text{ mi}
\]

Problem 19:

Right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \), \( AC = 4.5 \) mi, \( \angle B = 42^\circ \).

Step 1: \( \angle A \)

\[
\angle A = 90^\circ - 42^\circ = 48^\circ
\]

Step 2: Find \( BC \) (let \( BC = y \))

\( \tan(42^\circ) = \frac{AC}{BC} \implies y = \frac{4.5}{\tan(42^\circ)} \)
\( \tan(42^\circ) \approx 0.9004 \):
\[
y \approx \frac{4.5}{0.9004} \approx 5.0 \text{ mi}
\]

Step 3: Find \( AB \) (hypotenuse, \( z \))

\( \cos(42^\circ) = \frac{BC}{AB} \implies AB = \frac{BC}{\cos(42^\circ)} \), or \( \sin(42^\circ) = \frac{AC}{AB} \implies AB = \frac{4.5}{\sin(42^\circ)} \)
\( \sin(42^\circ) \approx 0.6691 \):
\[
AB \approx \frac{4.5}{0.6691} \approx 6.7 \text{ mi}
\]

Problem 21:

Right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \), \( AC = 29.3 \) mi, \( \angle A = 28^\circ \).

Step 1: \( \angle B \)

\[
\angle B = 90^\circ - 28^\circ = 62^\circ
\]

Step 2: Find \( BC \) (let \( BC = y \))

\( \tan(28^\circ) = \frac{BC}{AC} \implies y = 29.3 \times \tan(28^\circ) \)
\( \tan(28^\circ) \approx 0.5317 \):
\[
y \approx 29.3 \times 0.5317 \approx 15.6 \text{ mi}
\]

Step 3: Find \( AB \) (hypotenuse, \( z \))

\( \cos(28^\circ) = \frac{AC}{AB} \implies AB = \frac{29.3}{\cos(28^\circ)} \)
\( \cos(28^\circ) \approx 0.8829 \):
\[
AB \approx \frac{29.3}{0.8829} \approx 33.2 \text{ mi}
\]

Problem 23:

Right triangle \( \triangle ABC \) with \( \angle C = 90^\circ \), hypotenuse \( AB = 3 \) cm, \( \angle A = 40^\circ \).

Step 1: Find \( AC \) (let \( AC = x \))

\( \cos(40^\circ) = \frac{AC}{AB} \implies x = 3 \times \cos(40^\circ) \)
\( \cos(40^\circ) \approx 0.7660 \):
\[
x \approx 3 \times 0.7660 \approx 2.3 \text{ cm}
\]

Step 2: Find \( BC \) (let \( BC = y \))

\( \sin(40^\circ) = \frac{BC}{AB} \implies y = 3 \times \sin(40^\circ) \)
\( \sin(40^\circ) \approx 0.6428 \):
\[
y \approx 3 \times 0.6428 \approx 1.9 \text{ cm}
\]

Step 3: Find \( \angle B \)

\[
\angle B = 90^\circ - 40^\circ = 50^\circ
\]

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Final Answers (rounded as needed):

• Problem 13: \( AC \approx \boldsymbol{4.6} \) (units not specified, likely same as \( BC \))
• Problem 15: \( AC \approx \boldsymbol{6.2} \)
• Problem 17: \( \angle A = 28^\circ \), \( BC \approx 12.0 \) mi, \( AB \approx 25.6 \) mi
• Problem 19: \( \angle A = 48^\circ \), \( BC \approx 5.0 \) mi, \( AB \approx 6.7 \) mi
• Problem 21: \( \angle B = 62^\circ \), \( BC \approx 15.6 \) mi, \( AB \approx 33.2 \) mi
• Problem 23: \( AC \approx 2.3 \) cm, \( BC \approx 1.9 \) cm, \( \angle B = 50^\circ \)