Notes for Lesson 10 – 6: Trigonometric Ratios

10-6.1 – Finding Trigonometric Ratios

We learned that the 3 sides of a right triangle has a special relationship. Using the Pythagorean theorem, we could determine that a2+b2=c2 and therefore could find the length of any side given the other two. The angles of a right triangle also have certain relationships to the sides. These are called the trigonometric ratios. There are three ratios we will be looking at.

They are:

S ine(sin)= Lengthof the opposite legLengthof the hypotenuse

Cosine (cos)¿ Lenght of adjacent legLength of the hypotenuse

Tangent ( tan )= Lengthof the opposite legLength of the adjacent leg

Examples:

sin A=¿ 1517

¿ sin B= 817

sin F=1215

=45

sin E= 915

=35

cos A= 817

cos B=1517

cos F= 915

=35

cos E=1215

=45

tan A=158

tanB= 815

tan F=129

=43

tan E= 912

= 34

10-6.2 – Finding the ratio and angle using the calculator or chart

Show them how to read the chart and use the calculator to find the ratios and to find the angle given a ratio

Examples (make sure calculators are in degrees):

Sin (80) = .9848 Cos (15) = .9659 Tan (45) = 1.0000 Sin (9) = .1564

sin−1 ( .7660 )=50° cos−1 ( .8829 )=28 ° tan−1 (.6609 )=31° cos−1(.2588)=75 °

10-6.3 – Finding the missing side length

To find the missing side length set up the formula for a trigonometric ratio that match the information you have.

Examples:

Since you have the leg opposite the angle You have the adjacent leg and the hypotenuse you would use the and the hypotenuse so usesin ratio. The cos

sin (48 )= x14

cos (35 )=1.575x

14∗sin (48 )=x x∗cos (35 )=1.575

14∗.7431=x x= 1.575cos (35)

x=10.4034 x=1.575.8192

x=1.9226

10-6.4 – Finding the missing angle.

Finding the missing angle is the same as a missing side. You just divide the ratio out and find the angle associated with that ratio.

Examples:

cos ( x )= 510

tan ( x )= 56.7

cos ( x )=0.5 tan ( x )=.7463x=60 ° x≈36.7 °

10-6.5 – Using an angle of Elevation or Depression

If a kite has a 60 meter string on it and the angle of elevation is 65 °. How high is the kite if you are holding it 2 meters off the ground?

Use the sin since you have the opposite leg and the hypotenuse

60 sin 65 °= x60

X 60∗sin 65 °=x 65 60∗.9063=x

54.4=x plus the2metersis 56.4meters

A pilot is flying at 30,000 feet. The angle of depression from the plane to the start of the runway is 8 °. How far from the runway is the plane when it starts its decent.

8 °

30,000

X

The angle inside the triangle will be 82 ° so you have the opposite and the adjacent so use the tangent.

tan82 °= x30000

30000∗tan 82 °=x30000∗7.1154=x213,462 ft∨40.43miles=x