Solving for Unknown Side Lengths Using Sine or Cosine

Lesson 2: The Sine and Cosine Ratios - Solving for Unknown Side Lengths Using Sine or Cosine

   Constructing Knowledge

Recall the steps used to solve for the side length or the angle measure when using the tangent ratio from Lesson 1.

Step 1: Identify and label the sides as being adjacent to, opposite or the hypotenuse in relation to the angle indicated.

Step 2: State the appropriate ratio.

Step 3: Substitute known values and calculate the unknown value.

   Multimedia

A video demonstration of the solution for Example 1 has been provided.

EXAMPLE 1

---

Determine the length of x, to the nearest tenth.

Solution

Step 1: Identify and label the sides as being adjacent to, opposite or the hypotenuse in relation to the angle indicated.



Step 2: State the appropriate ratio.

The length adjacent to 34° and the hypotenuse are involved, so use the cosine ratio.

\(\text{cos}\,\theta=\frac{\text{length adjacent to}\,\theta}{\text{hypotenuse}}\)

Step 3: Substitute known values and calculate the unknown value.

\(\begin{align} \text{cos}\,\theta&=\frac{\text{adj}}{\text{adj}} \\ \\ \text{cos}\,34°&=\frac{x}{13\,\text{ft}} \\ \\ {\color{red}{13\,\text{ft}}}\times \text{cos}\,34°&=\frac{x}{\cancel{13\,\text{ft}}}\times \cancel{\color{red}{13\,\text{ft}}} \\ \\ 10.777...\,\text{ft}&=x \\ \\ 10.8\,\text{ft}&=x \\ \end{align}\)

The length of x is approximately 10.8 ft.

EXAMPLE 2

---

Determine the length of AC, to the nearest tenth.

Solution

Step 1: Identify and label the sides as being adjacent to, opposite or the hypotenuse in relation to the angle indicated.



Step 2: State the appropriate ratio.

The length opposite of 20° and the hypotenuse are involved, so use the sine ratio.

\(\text{sin}\,\theta=\frac{\text{length opposite to}\,\theta}{\text{hypotenuse}}\)

Step 3: Substitute known values and calculate the unknown value.

\(\begin{align} \text{sin}\,\theta&=\frac{\text{opp}}{\text{hyp}} \\ \\ \text{sin}\,20°&=\frac{13\,\text{m}}{\text{AC}} \\ \\ {\color{red}{\text{AC}}}\times \text{sin}\,20°&=\frac{13\,\text{m}}{\cancel{\text{AC}}}\times \cancel{\color{red}{\text{AC}}} \\ \\ \text{AC}\times \text{sin}\,20°&=13\,\text{m} \\ \\ \frac{\text{AC}\times \cancel{\text{sin}\,20°}}{\cancel{\color{red}{\text{sin}\,20°}}}&=\frac{13\,\text{m}}{\color{red}{\text{sin}\,20°}} \\ \\ \text{AC}&=38\,\text{m} \\ \end{align}\)

The length of AC is approximately 38.0 m.




2014 © Alberta Distance Learning Centre