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Trigonometic Ratios (SOH CAH TOA): Word Problems (With Clues ... - Free Printable

Trigonometic Ratios (SOH CAH TOA): Word Problems (With Clues ...

Educational worksheet: Trigonometic Ratios (SOH CAH TOA): Word Problems (With Clues .... Download and print for classroom or home learning activities.

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Show Answer Key & Explanations Step-by-step solution for: Trigonometic Ratios (SOH CAH TOA): Word Problems (With Clues ...
Let's solve each problem step by step.

---

Problem 1:


ABC is a right triangle. AB = 7 cm, angle ABC = 90°, and angle ACB = 64°. Calculate the length of BC.

#### Solution:
- In a right triangle, we can use trigonometric ratios to find the missing side.
- Here, AB is the opposite side to angle ACB, and BC is the adjacent side to angle ACB.
- We use the tangent function: \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\).
- Given: \(\tan(64^\circ) = \frac{AB}{BC} = \frac{7}{BC}\).
- Solving for \(BC\):
\[
BC = \frac{7}{\tan(64^\circ)}
\]
- Using a calculator:
\[
\tan(64^\circ) \approx 2.0503
\]
\[
BC = \frac{7}{2.0503} \approx 3.41 \text{ cm}
\]

#### Final Answer:
\[
\boxed{3.41}
\]

---

Problem 2:


The lengths of the sides of a right triangle are 5 cm, 12 cm, and 13 cm. Calculate the size of the other two angles of this triangle.

#### Solution:
- In a right triangle, one angle is always 90°. The other two angles are complementary (sum to 90°).
- Let's denote the angles as \(\theta\) and \(90^\circ - \theta\).
- We can use the sine or cosine function to find one of the angles. For example, using the sine function:
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]
- Here, the hypotenuse is 13 cm, and one of the legs is 5 cm (opposite to \(\theta\)):
\[
\sin(\theta) = \frac{5}{13}
\]
- Solving for \(\theta\):
\[
\theta = \arcsin\left(\frac{5}{13}\right)
\]
- Using a calculator:
\[
\theta \approx \arcsin(0.3846) \approx 22.62^\circ
\]
- The other angle is:
\[
90^\circ - \theta \approx 90^\circ - 22.62^\circ \approx 67.38^\circ
\]

#### Final Answer:
\[
\boxed{22.6, 67.4}
\]

---

Problem 3:


The perimeter of a right triangle is 24 cm. The length of one of the sides is 10 cm, the length of the other side is 8 cm. Calculate the size of the smallest angle of this triangle.

#### Solution:
- The perimeter of the triangle is the sum of its sides:
\[
\text{Perimeter} = 10 + 8 + \text{hypotenuse} = 24
\]
- Solving for the hypotenuse:
\[
10 + 8 + \text{hypotenuse} = 24
\]
\[
\text{hypotenuse} = 24 - 18 = 6 \text{ cm}
\]
- Now, we have the sides of the triangle: 6 cm, 8 cm, and 10 cm. This is a right triangle because \(6^2 + 8^2 = 10^2\).
- To find the smallest angle, we use the smallest side (6 cm) as the opposite side to the smallest angle. Let the smallest angle be \(\theta\):
\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{10} = 0.6
\]
- Solving for \(\theta\):
\[
\theta = \arcsin(0.6)
\]
- Using a calculator:
\[
\theta \approx \arcsin(0.6) \approx 36.87^\circ
\]

#### Final Answer:
\[
\boxed{36.9}
\]

---

Problem 4:


A rectangle has a length of 14 cm and width of 8 cm. What angle does the diagonal make with the longest side?

#### Solution:
- The diagonal of the rectangle forms a right triangle with the length and width of the rectangle.
- Let the angle between the diagonal and the longest side (length = 14 cm) be \(\theta\).
- We use the tangent function:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{width}}{\text{length}} = \frac{8}{14} = \frac{4}{7}
\]
- Solving for \(\theta\):
\[
\theta = \arctan\left(\frac{4}{7}\right)
\]
- Using a calculator:
\[
\theta \approx \arctan(0.5714) \approx 29.74^\circ
\]

#### Final Answer:
\[
\boxed{29.7}
\]

---

Problem 5:


A ladder is 5 m long. The ladder rests against a vertical wall. The foot of the ladder rests on horizontal ground. The ladder makes an angle of 82° with the ground. How far up the wall does the ladder reach?

#### Solution:
- The ladder forms a right triangle with the ground and the wall.
- The length of the ladder is the hypotenuse (5 m).
- The height the ladder reaches up the wall is the opposite side to the angle of 82°.
- We use the sine function:
\[
\sin(82^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\text{height}}{5}
\]
- Solving for the height:
\[
\text{height} = 5 \cdot \sin(82^\circ)
\]
- Using a calculator:
\[
\sin(82^\circ) \approx 0.9903
\]
\[
\text{height} = 5 \cdot 0.9903 \approx 4.95 \text{ m}
\]

#### Final Answer:
\[
\boxed{4.95}
\]

---

Final Answers:


1. \(\boxed{3.41}\)
2. \(\boxed{22.6, 67.4}\)
3. \(\boxed{36.9}\)
4. \(\boxed{29.7}\)
5. \(\boxed{4.95}\)
Parent Tip: Review the logic above to help your child master the concept of right triangle word problems worksheet.
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