Question

The sides of a triangle are 17 ft., 19 ft., and 15 ft. Find the measure of the angle opposite the longest side. Round your answer to the nearest degree.
(1 point)
Responses

107°
107°

73°
73°

59°
59°

49°

Find all solutions of the equation 2 sine x cosine 2 x minus cosine 2 x equals 0 over the interval 0 less-than x less-than-or-equal-to pi.
(1 point)
Responses

x equals start fraction pi over 6 end fraction comma start fraction pi over 2 end fraction comma start fraction 5 pi over 6 end fraction comma pi
Image with alt text: x equals start fraction pi over 6 end fraction comma start fraction pi over 2 end fraction comma start fraction 5 pi over 6 end fraction comma pi

x equals start fraction pi over 6 end fraction comma start fraction pi over 4 end fraction comma start fraction 3 pi over 4 end fraction comma start fraction 5 pi over 6 end fraction
Image with alt text: x equals start fraction pi over 6 end fraction comma start fraction pi over 4 end fraction comma start fraction 3 pi over 4 end fraction comma start fraction 5 pi over 6 end fraction

x equals start fraction pi over 4 end fraction comma start fraction pi over 3 end fraction comma start fraction 2 pi over 3 end fraction comma start fraction 3 pi over 4 end fraction
Image with alt text: x equals start fraction pi over 4 end fraction comma start fraction pi over 3 end fraction comma start fraction 2 pi over 3 end fraction comma start fraction 3 pi over 4 end fraction

x equals start fraction pi over 3 end fraction comma start fraction pi over 2 end fraction comma start fraction 2 pi over 3 end fraction comma pi

Answer 1

To solve the problems one by one:

### Problem 1: Find the measure of the angle opposite the longest side.

1. The sides of the triangle are 17 ft, 19 ft, and 15 ft. The longest side is 19 ft.
2. We can use the Law of Cosines to find the angle opposite the longest side:

\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]

Here, let \(c = 19\), \(a = 17\), and \(b = 15\).

Plugging in the values gives:

\[
19^2 = 17^2 + 15^2 - 2 \cdot 17 \cdot 15 \cdot \cos(C)
\]

\[
361 = 289 + 225 - 510 \cdot \cos(C)
\]

\[
361 = 514 - 510 \cdot \cos(C)
\]

\[
510 \cdot \cos(C) = 514 - 361
\]

\[
510 \cdot \cos(C) = 153
\]

\[
\cos(C) = \frac{153}{510} = 0.3
\]

Now find \(C\):

\[
C = \cos^{-1}(0.3)
\]

Using a calculator:

\[
C \approx 72.34^\circ
\]

Rounding this to the nearest degree gives \(73^\circ\).

**Answer**: 73°

---

### Problem 2: Solve the equation \(2 \sin(x) \cos(2x) - \cos(2x) = 0\) over the interval \(0 < x \leq \pi\).

1. Factor the equation:

\[
\cos(2x)(2\sin(x) - 1) = 0
\]

This gives us two equations to solve:

- \( \cos(2x) = 0 \)
- \( 2\sin(x) - 1 = 0 \)

#### Solving \( \cos(2x) = 0 \):

\[
2x = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z})
\]

So,

\[
x = \frac{\pi}{4} + \frac{n\pi}{2}
\]

For \(n = 0\):

\[
x = \frac{\pi}{4}
\]

For \(n = 1\):

\[
x = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}
\]

For \(n = 2\):

\[
x = \frac{\pi}{4} + \pi = \frac{5\pi}{4} \text{ (not in the interval)}
\]

Contributions from \( \cos(2x) = 0 \): \(x = \frac{\pi}{4}, \frac{3\pi}{4}\)

#### Solving \(2\sin(x) - 1 = 0\):

\[
\sin(x) = \frac{1}{2}
\]

This occurs at:

\[
x = \frac{\pi}{6}, \frac{5\pi}{6} \quad (0 < x \leq \pi)
\]

### Final Solutions:
Combining our results:

- From \( \cos(2x) = 0\): \(x = \frac{\pi}{4}, \frac{3\pi}{4}\)
- From \(2\sin(x) - 1 = 0\): \(x = \frac{\pi}{6}, \frac{5\pi}{6}\)

**All solutions**: \(x = \frac{\pi}{6}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{6}\)

**Answer**: \(x = \frac{\pi}{6}, \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{6}\)