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# 5.4: Sine Ratio

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Review the different parts of right triangles.
• Identify and use the sine ratio in a right triangle.

## Review: Parts of a Triangle

The sine and cosine ratios relate opposite and adjacent sides to the hypotenuse. You already learned these terms in the previous lesson, but they are important to review and commit to memory.

The hypotenuse of a triangle is always opposite the right angle and is the longest side of a right triangle.

The terms adjacent and opposite depend on which angle you are referencing:

A side adjacent to an angle is the leg of the triangle that helps form the angle.

A side opposite to an angle is the leg of the triangle that does not help form the angle.

The hypotenuse is _____________________________________________________.

The opposite side is ____________________________________________________.

Example 1

Examine the triangle in the diagram below.

Identify which leg is adjacent to angle N\begin{align*}N\end{align*}, which leg is opposite to angle N\begin{align*}N\end{align*}, and which segment is the hypotenuse.

The first part of the question asks you to identify the leg adjacent to N\begin{align*}\angle{N}\end{align*}. Since an adjacent leg is the one that helps to form the angle and is not the hypotenuse, it must be MN¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{MN}\end{align*}.

The next part of the question asks you to identify the leg opposite N\begin{align*}\angle{N}\end{align*}. Since an opposite leg is the leg that does not help to form the angle, it must be LM¯¯¯¯¯¯¯¯¯\begin{align*}\overline{LM}\end{align*}.

The hypotenuse is always opposite the right angle, so in this triangle it is segment LN¯¯¯¯¯¯¯¯\begin{align*}\overline{LN}\end{align*}.

1. Which side of a right triangle is the longest side? _____________________________

2. Describe where the side you answered in #1 above is in relation to the right angle:

\begin{align*}\; \;\end{align*}

\begin{align*}\; \;\end{align*}

3. Which side of a right triangle does not help to make the right angle? _____________________________

4. Which side of a right triangle helps to make the right angle and is NOT the hypotenuse? _____________________________

## The Sine Ratio

Another important trigonometric ratio is sine. A sine ratio must always refer to a particular angle in a right triangle. The sine of an angle is the ratio of the length of the leg opposite the angle to the length of the hypotenuse.

This means that the sine ratio is: the ____________________ side divided by the _______________________.

Remember that in a ratio, you list the first item on top of the fraction and the second item on the bottom. So, the ratio of the sine will be:

sinθ=oppositehypotenuse\begin{align*} \sin \theta = \frac{opposite}{hypotenuse}\end{align*}

Example 2

What are sinA\begin{align*}\sin A\end{align*} and sinB\begin{align*}\sin B\end{align*} in the triangle below?

To find the solutions, you must identify the sides you need and build the ratios carefully. In the sine ratio, we will need the opposite side and the hypotenuse.

Remember, the hypotenuse of a right triangle is across from the right angle. The opposite side depends on which angle we are using.

The hypotenuse is the segment ___________, which is _______ cm long.

For angle A\begin{align*}A\end{align*}:

The side opposite angle A\begin{align*}A\end{align*} is the segment ___________, which is _______ cm long.

For angle B\begin{align*}B\end{align*}:

The side opposite angle B\begin{align*}B\end{align*} is the segment ___________, which is _______ cm long.

sinAsinB=oppositehypotenuse=35=oppositehypotenuse=45\begin{align*}\sin A & = \frac {opposite}{hypotenuse} = \frac {3}{5}\\ \sin B & = \frac {opposite}{hypotenuse} = \frac {4}{5}\end{align*}

So, \begin{align*}\sin A = \frac {3}{5}\end{align*} and \begin{align*}\sin B = \frac {4}{5}\end{align*}.

Your friend did the following problem and asked you if it was correct:

Find \begin{align*}\sin X\end{align*} using the triangle below.

\begin{align*}\sin X = \frac{opposite}{hypotenuse} = \frac{5}{12}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;} \end{align*}

2. If not, where is the mistake in the problem? Describe the mistake in words and explain to your friend how she should have done the problem correctly.

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

\begin{align*}{\;}\end{align*}

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