Trigonometric Ratios SOHCAHTOA worksheet with problems on finding sine, cosine, tangent, and solving for missing sides and angles in right triangles.

Problem Analysis and Solution

The worksheet involves trigonometric ratios and their applications. Let's solve each part step by step.

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#### Part 1: Find the trigonometric ratios for the given right triangle

The triangle is a right triangle with:
- Hypotenuse \( AC = 14 \)
- Opposite side to angle \( A \): \( BC = 7 \)
- Adjacent side to angle \( A \): \( AB = 7\sqrt{3} \)

We need to find:
1. \( \sin A \)
2. \( \cos A \)
3. \( \tan A \)

Using the SOHCAHTOA mnemonic:
- \( \sin A = \frac{\text{opposite}}{\text{hypotenuse}} \)
- \( \cos A = \frac{\text{adjacent}}{\text{hypotenuse}} \)
- \( \tan A = \frac{\text{opposite}}{\text{adjacent}} \)

##### Step 1: Calculate \( \sin A \)
\[
\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{14} = \frac{1}{2}
\]

##### Step 2: Calculate \( \cos A \)
\[
\cos A = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{7\sqrt{3}}{14} = \frac{\sqrt{3}}{2}
\]

##### Step 3: Calculate \( \tan A \)
\[
\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB} = \frac{7}{7\sqrt{3}} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}
\]

##### Final Answers for Part 1:
\[
\sin A = \frac{1}{2}, \quad \cos A = \frac{\sqrt{3}}{2}, \quad \tan A = \frac{\sqrt{3}}{3}
\]

---

#### Part 2: Find the value of each expression

We need to evaluate:
1. \( \sin 30^\circ \)
2. \( \cos 45^\circ \)
3. \( \tan 45^\circ \)

##### Step 1: Evaluate \( \sin 30^\circ \)
From trigonometric values:
\[
\sin 30^\circ = \frac{1}{2}
\]

##### Step 2: Evaluate \( \cos 45^\circ \)
From trigonometric values:
\[
\cos 45^\circ = \frac{\sqrt{2}}{2}
\]

##### Step 3: Evaluate \( \tan 45^\circ \)
From trigonometric values:
\[
\tan 45^\circ = 1
\]

##### Final Answers for Part 2:
\[
\sin 30^\circ = \frac{1}{2}, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}, \quad \tan 45^\circ = 1
\]

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#### Part 3: Use trigonometric ratios to find the missing variables

##### Subpart (a):
The triangle has:
- One angle \( 45^\circ \)
- Hypotenuse \( 16 \)
- Two legs \( x \) and \( y \)

Since the triangle is a 45-45-90 triangle, the legs are equal, and the relationship between the hypotenuse and a leg is:
\[
\text{Leg} = \frac{\text{Hypotenuse}}{\sqrt{2}}
\]

##### Step 1: Calculate \( x \) and \( y \)
\[
x = y = \frac{16}{\sqrt{2}} = \frac{16\sqrt{2}}{2} = 8\sqrt{2}
\]

Approximating \( \sqrt{2} \approx 1.414 \):
\[
x = y = 8 \times 1.414 \approx 11.3
\]

##### Final Answers for Subpart (a):
\[
x = 11.3, \quad y = 11.3
\]

##### Subpart (b):
The triangle has:
- Hypotenuse \( 13 \)
- One angle \( 60^\circ \)
- One angle \( z^\circ \)
- One leg \( y \)

##### Step 1: Determine \( z^\circ \)
The sum of angles in a triangle is \( 180^\circ \). Since one angle is \( 90^\circ \) (right angle) and another is \( 60^\circ \):
\[
z^\circ = 180^\circ - 90^\circ - 60^\circ = 30^\circ
\]

##### Step 2: Calculate \( y \)
Using the sine function for the angle \( 60^\circ \):
\[
\sin 60^\circ = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{13}
\]
\[
\sin 60^\circ = \frac{\sqrt{3}}{2}
\]
\[
\frac{\sqrt{3}}{2} = \frac{y}{13}
\]
\[
y = 13 \cdot \frac{\sqrt{3}}{2} = \frac{13\sqrt{3}}{2}
\]

Approximating \( \sqrt{3} \approx 1.732 \):
\[
y = \frac{13 \times 1.732}{2} \approx \frac{22.516}{2} \approx 11.3
\]

##### Final Answers for Subpart (b):
\[
z^\circ = 30^\circ, \quad y = 11.3
\]

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Final Answer:

\[
\boxed{
\begin{array}{l}
\text{Part 1: } \sin A = \frac{1}{2}, \cos A = \frac{\sqrt{3}}{2}, \tan A = \frac{\sqrt{3}}{3} \\
\text{Part 2: } \sin 30^\circ = \frac{1}{2}, \cos 45^\circ = \frac{\sqrt{2}}{2}, \tan 45^\circ = 1 \\
\text{Part 3(a): } x = 11.3, y = 11.3 \\
\text{Part 3(b): } z^\circ = 30^\circ, y = 11.3
\end{array}
}
\]