Trig Ratios Sum Em Activity | Teorema de pitágoras, Matemática ... - Free Printable

The image contains a set of eight problems involving right triangles, where the goal is to find the value of \( x \) using trigonometric ratios (sine, cosine, or tangent). Each problem requires rounding the answer to the nearest tenth. Below, I will solve each problem step by step.

---

Problem 1:

- Given: A right triangle with one angle \( 70^\circ \), the adjacent side \( 36 \), and the hypotenuse \( x \).
- To find: The value of \( x \).

#### Solution:
We use the cosine function:
\[
\cos(70^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{36}{x}
\]
Rearranging for \( x \):
\[
x = \frac{36}{\cos(70^\circ)}
\]
Using a calculator:
\[
\cos(70^\circ) \approx 0.342
\]
\[
x = \frac{36}{0.342} \approx 105.26
\]
Rounding to the nearest tenth:
\[
x \approx 105.3
\]

Answer: \( x \approx 105.3 \)

---

Problem 2:

- Given: A right triangle with one angle \( 31^\circ \), the opposite side \( 18 \), and the hypotenuse \( x \).
- To find: The value of \( x \).

#### Solution:
We use the sine function:
\[
\sin(31^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{18}{x}
\]
Rearranging for \( x \):
\[
x = \frac{18}{\sin(31^\circ)}
\]
Using a calculator:
\[
\sin(31^\circ) \approx 0.515
\]
\[
x = \frac{18}{0.515} \approx 34.97
\]
Rounding to the nearest tenth:
\[
x \approx 35.0
\]

Answer: \( x \approx 35.0 \)

---

Problem 3:

- Given: A right triangle with one angle \( 85^\circ \), the adjacent side \( 20 \), and the opposite side \( x \).
- To find: The value of \( x \).

#### Solution:
We use the tangent function:
\[
\tan(85^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{20}
\]
Rearranging for \( x \):
\[
x = 20 \cdot \tan(85^\circ)
\]
Using a calculator:
\[
\tan(85^\circ) \approx 11.43
\]
\[
x = 20 \cdot 11.43 \approx 228.6
\]
Rounding to the nearest tenth:
\[
x \approx 228.6
\]

Answer: \( x \approx 228.6 \)

---

Problem 4:

- Given: A right triangle with one angle \( 21^\circ \), the adjacent side \( 20 \), and the opposite side \( x \).
- To find: The value of \( x \).

#### Solution:
We use the tangent function:
\[
\tan(21^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{20}
\]
Rearranging for \( x \):
\[
x = 20 \cdot \tan(21^\circ)
\]
Using a calculator:
\[
\tan(21^\circ) \approx 0.384
\]
\[
x = 20 \cdot 0.384 \approx 7.68
\]
Rounding to the nearest tenth:
\[
x \approx 7.7
\]

Answer: \( x \approx 7.7 \)

---

Problem 5:

- Given: A right triangle with one angle \( 37^\circ \), the hypotenuse \( 17 \), and the opposite side \( x \).
- To find: The value of \( x \).

#### Solution:
We use the sine function:
\[
\sin(37^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{17}
\]
Rearranging for \( x \):
\[
x = 17 \cdot \sin(37^\circ)
\]
Using a calculator:
\[
\sin(37^\circ) \approx 0.602
\]
\[
x = 17 \cdot 0.602 \approx 10.23
\]
Rounding to the nearest tenth:
\[
x \approx 10.2
\]

Answer: \( x \approx 10.2 \)

---

Problem 6:

- Given: A right triangle with one angle \( 30^\circ \), the adjacent side \( 20 \), and the hypotenuse \( x \).
- To find: The value of \( x \).

#### Solution:
We use the cosine function:
\[
\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{20}{x}
\]
Rearranging for \( x \):
\[
x = \frac{20}{\cos(30^\circ)}
\]
Using a calculator:
\[
\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866
\]
\[
x = \frac{20}{0.866} \approx 23.09
\]
Rounding to the nearest tenth:
\[
x \approx 23.1
\]

Answer: \( x \approx 23.1 \)

---

Problem 7:

- Given: A right triangle with one angle \( 15^\circ \), the hypotenuse \( 48 \), and the opposite side \( x \).
- To find: The value of \( x \).

#### Solution:
We use the sine function:
\[
\sin(15^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{48}
\]
Rearranging for \( x \):
\[
x = 48 \cdot \sin(15^\circ)
\]
Using a calculator:
\[
\sin(15^\circ) \approx 0.259
\]
\[
x = 48 \cdot 0.259 \approx 12.43
\]
Rounding to the nearest tenth:
\[
x \approx 12.4
\]

Answer: \( x \approx 12.4 \)

---

Problem 8:

- Given: A right triangle with one angle \( 40^\circ \), the adjacent side \( 125 \), and the hypotenuse \( x \).
- To find: The value of \( x \).

#### Solution:
We use the cosine function:
\[
\cos(40^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{125}{x}
\]
Rearranging for \( x \):
\[
x = \frac{125}{\cos(40^\circ)}
\]
Using a calculator:
\[
\cos(40^\circ) \approx 0.766
\]
\[
x = \frac{125}{0.766} \approx 163.24
\]
Rounding to the nearest tenth:
\[
x \approx 163.2
\]

Answer: \( x \approx 163.2 \)

---

Final Answers:

\[
\boxed{
\begin{array}{ll}
1. & x \approx 105.3 \\
2. & x \approx 35.0 \\
3. & x \approx 228.6 \\
4. & x \approx 7.7 \\
5. & x \approx 10.2 \\
6. & x \approx 23.1 \\
7. & x \approx 12.4 \\
8. & x \approx 163.2 \\
\end{array}
}
\]