Jeremy was in the park flying the kite. It was a windy day and the kite was airborne within seconds. The kite was flying at the end of a string 48 meters long, making an angle of 75° with the ground. How can Jeremy figure out the altitude of his kite?

In this concept, you will learn to determine and use the Sine ratio.

### Sine Ratio

You have used the TI-calculator to determine the measure of an angle using the inverse sine function

when value of the ratio was known. If then using the TI-calculator displayed the measure of 51.7° for the angle.

The TI-calculator can also be used to find the ratio when the measure of the angle is known. If the measure of

then the value of the ratio can be found by following the Key Press History:

On the calculator screen the following is displayed:

This decimal should always be rounded to the nearest ten thousandth (four places after the decimal).

This sine ratio for an angle of 48° will remain constant regardless of the size of the triangle. The sides will always be in the same proportion to each other.

The value of the Sine ratio will always be between zero and one.

The Sine ratio is related to one of the acute angles of a right triangle such that the sine of the acute angle is the ratio of the side opposite the specific acute angle (the reference angle) to the hypotenuse. The Sine ratio can be written as . This equation has three parts to it – an angle and two sides. When the lengths of the two sides are known, the measure of the angle can be calculated. When the measure of an angle and the length of one side is known, the length of the other side can be calculated.

It is the sides of the triangle that determine the trigonometric ratio that will be used to calculate the measure of an angle or the length of a side. To use the sine ratio, the opposite side and the hypotenuse of the right triangle must be indicated. These two sides will have values on them if the measure of an angle is to be calculated using the sine ratio. If the length of a side is to be calculated using the sine ratio then one of the sides will display a value and the other will display a variable (the side that is unknown).

Let’s look at the following right triangle to see how this works.

The measure of

is 65°. The length of the hypotenuse is 28 meters. The side has the variable ‘ ’ on it which means this is the side that is unknown and its length must be calculated.The sides of the triangle can be named using the acute angle

which is the reference angle for this triangle.

The two sides that are indicated on this triangle are the hypotenuse and the opposite. The Sine ratio is the ratio of the opposite side to the hypotenuse.

First, write the sine ratio using words.

\begin{align*}\sin B = \frac{\text{opposite}}{\text{hypotenuse}}\end{align*}

Next, write the sine ratio using symbols.

\begin{align*}\sin B = \frac{AC}{AB}\end{align*}

Next, fill all known values into the equation.

.\begin{align*}\sin 65^\circ = \frac{X}{28}\end{align*}

Next, use the TI-calculator to find the value of

. Round the decimal to the nearest ten thousandth.

Next, substitute this value into the equation.

\begin{align*}\begin{array}{rcl} \sin 65^\circ &=& \frac{X}{28} \\ 0.9063 &=& \frac{X}{28} \end{array}\end{align*}

Next, multiply both sides of the equation by 28 to solve for the variable.

\begin{align*}\begin{array}{rcl} 0.9063 &=& \frac{X}{28} \\ 28(0.9063) &=& 28 \left( \frac{X}{28} \right) \\ 25.3764 &=& { \overset{1} {\cancel{28}} } \left ( \frac{X}{ \cancel{28}} \right) \\ 25.38 &=& X \end{array}\end{align*}

The length of the opposite side of the right triangle is 25.38 meters.

When calculating the length of a side of a right triangle, the answer is usually rounded to the nearest hundredth unless otherwise stated.

When calculating the measure of an angle of a right triangle, the answer is usually rounded to the nearest tenth unless otherwise stated.

### Examples

#### Example 1

Earlier, you were given a problem about Jeremy and his kite. He needs to figure out how high his kite is above the ground.

Jeremy can use the sine ratio to calculate the answer.

First, draw and label a right triangle to model the kite flying.

First, using the reference angle

, name the sides of the right triangle.

First, write the sine ratio using words.

\begin{align*}\sin A = \frac{\text{opposite}}{\text{hypotenuse}}\end{align*}

Next, write the sine ratio using symbols.

\begin{align*}\sin A = \frac{BC}{AB}\end{align*}

Next, fill all known values into the equation.

.

Next, use the TI-calculator to find the value of

.\begin{align*}\sin 75^\circ =0.9659\end{align*}

Next, substitute this value into the equation.

\begin{align*}\begin{array}{rcl} \sin 75^\circ &=& \frac{X}{48} \\ 0.9659 &=& \frac{X}{48} \end{array}\end{align*}

Next, multiply both sides of the equation by 48 to solve for the variable.

\begin{align*}\begin{array}{rcl} 0.9659 &=& \frac{X}{48} \\ 48(0.9659) &=& 48 \left( \frac{X}{48} \right) \\ 46.36 &=& { \overset{1} {\cancel{48}} } \left ( \frac{X}{ \cancel{48}} \right) \\ 46.36 &=& X \end{array}\end{align*}

The answer is 46.36.

Jeremy’s kite is 46.36 meters above the ground.

#### Example 2

For the following right triangle calculate the length of side ‘

’ to the nearest hundredth.

First, write down what you know from the right triangle.

\begin{align*}\begin{array}{rcl} \angle F &=& 30^\circ \\ \overline{DE} &=& 17.05 \ \text{units} \\ \overline{EF} &=& X \end{array}\end{align*}

Next, use the reference angle

to name the sides of the triangle.

The two sides that are indicated on this triangle are the hypotenuse and the opposite. The Sine ratio is the ratio of the opposite side to the hypotenuse.

First, write the sine ratio using words.

\begin{align*}\sin F = \frac{\text{opposite}}{\text{hypotenuse}} \end{align*}

Next, write the sine ratio using symbols.

Next, fill all known values into the equation.

.

Next, use the TI-calculator to find the value of

.

\begin{align*}\sin 30^\circ = 0.5\end{align*}

Next, substitute this value into the equation.

\begin{align*}\begin{array}{rcl} \sin 30^\circ &=& \frac{17.05}{X} \\ 0.5 &=& \frac{17.05}{X} \end{array}\end{align*}

Next, multiply both sides of the equation by

to clear the denominator.Then, divide both sides of the equation by 0.5 to solve for the variable.

\begin{align*}\begin{array}{rcl} 0.5 \ X &=& 17.05 \\ \frac{ { \overset{1} { \cancel{0.5}}} X }{ \cancel{0.5}} &=& \frac{17.05}{0.5} \\ X &=& 3.41 \end{array}\end{align*}

The answer is 34.10.

The length of the hypotenuse of the right triangle is 34.10 units.

#### Example 3

For the following right triangle calculate the length of side ‘\begin{align*}X\end{align*}’.

First, write down what you know from the right triangle.

\begin{align*}\begin{array}{rcl} \angle C &=& 32^\circ \\ \overline{BC} &=& 15 \ \text{feet} \\ \overline{AB} &=& X \end{array} \end{align*}

Next, use the reference angle

to name the sides of the triangle.

The two sides that are indicated on this triangle are the hypotenuse and the opposite. The Sine ratio is the ratio of the opposite side to the hypotenuse.

First, write the sine ratio using words.

Next, write the sine ratio using symbols.

Next, fill all known values into the equation.

.\begin{align*}\sin 32^\circ = \frac{X}{15}\end{align*}

Next, use the TI-calculator to find the value of

.

\begin{align*}\sin 32=0.5299\end{align*}

Next, substitute this value into the equation.

\begin{align*}\begin{array}{rcl} \sin 32^\circ &=& \frac{X}{15} \\ 0.5299 &=& \frac{X}{15} \end{array}\end{align*}

Next, multiply both sides of the equation by 15 to solve for the variable.

\begin{align*} \begin{array}{rcl} 0.5299 &=& \frac{X}{15} \\ 15(0.5299) &=& 15 \left( \frac{X}{15} \right) \\ 15(0.5299) &=& { \overset{1} {\cancel{15}} } \left ( \frac{X}{ \cancel{15}} \right) \\ 7.9485 &=& X \\ 7.95 &=& X \end{array}\end{align*}

The answer is 7.95.

The length of the opposite side of the right triangle is 7.95 feet.

#### Example 4

For the following mathematical solution, using the sine ratio to determine the length of the hypotenuse of the right triangle, briefly tell what is happening in each part of the solution.

\begin{align*}\begin{array}{rcl} \sin 36^\circ &=& \frac{11}{X} \\ 0.5878 &=& \frac{11}{X} \\ X(0.5878) &=& X \left( \frac{11}{X} \right) \\ 0.5878 X &=& { \overset{1} {\cancel{X}} } \left ( \frac{11}{ \cancel{X}} \right) \\ 0.5878 X &=& 11 \\ \frac{ \cancel{0.5878} X}{\cancel{0.5878}} &=& \frac{11} {0.5878} \\ X &=& 18.71 \end{array}\end{align*}

The answer is 18.71 cm.

The length of the hypotenuse of the right triangle is 18.71 cm.

### Review

Use a calculator to find each Sine. You may round to the nearest hundredth.

1. Sine 55°

2. Sine 25°

3. Sine 11°

4. Sine 60°

5. Sine 75°

6. Sine 12°

7. Sine 29°

8. Sine 15°

Use the information given and what you have learned about trigonometric ratios to figure out the measure of each missing side. You may round when necessary.

9. Sine angle

, hypotenuse – 12, what is the length of the opposite side?10. Sine angle \begin{align*}E \ 65^\circ\end{align*}, hypotenuse – 8, what is the length of the opposite side?

11. Sine angle

, hypotenuse – 2, what is the length of the opposite side?12. Sine angle

, hypotenuse – 10, what is the length of the opposite side?13. Sine angle

, hypotenuse – 8, what is the length of the opposite side?14. Sine angle

, hypotenuse – 5, what is the length of the opposite side?15. Sine angle

, hypotenuse – 18, what is the length of the opposite side?### Review (Answers)

To see the Review answers, open this PDF file and look for section 7.16.