Pc Geometry Triangle Trig

Triangle Trigonometry

Topics ◆ Basis of Trigonometry ◆ The Six Ratios ◆ Solving Right Triangles ◆ Special Right Triangles ◆ Law of Sines ◆ Law of Cosines ©Carolyn C. Wheater, 2000

2

Basis of Trigonometry ◆Trigonometry, or "triangle measurement," developed as a means to calculate the lengths of sides of right triangles. ◆It is based upon similar triangle relationships. ©Carolyn C. Wheater, 2000

3

Right Triangle Trigonometry ◆ You can quickly prove that the two right triangles with an acute angle of 25°are similar ◆ All right triangles containing an angle of 25° are similar

You could think of this as the family of 25° right triangles. Every triangle in the family 25° is similar. We could imagine such a family of triangles for any acute angle.

©Carolyn C. Wheater, 2000

25°

4

Right Triangle Trigonometry ◆ In any right triangle in the family, the ratio of the side opposite the acute angle to the hypotenuse will always be the same, and the ratios of other pairs of sides will remain constant.

©Carolyn C. Wheater, 2000

5

The Six Ratios ◆ If the three sides of the right angle are labeled as us ten

po e

opposite

hy

the hypotenuse, ■ the side opposite a particular acute angle, A, and ■ the side adjacent to the acute angle A, ■

A

adjacent

◆ six different ratios are possible. ©Carolyn C. Wheater, 2000

6

The Six Ratios opposite sin( A) = hypotenuse

hypotenuse csc( A) = opposite

adjacent cos( A) = hypotenuse

hypotenuse sec( A) = adjacent

opposite tan( A) = adjacent

adjacent cot( A) = opposite

©Carolyn C. Wheater, 2000

7

Solving Right Triangles ◆ With these six ratios, it is possible to solve for any unknown side of the right triangle, if another side and an acute angle are known, or ■ to find the angle if two sides are known. ■

Once upon a time, students had to rely on tables to look up these values. Now the sine, cosine, and tangent of an angle can be found on your calculator. ©Carolyn C. Wheater, 2000

8

Sample Problem ◆ In right triangle ABC, hypotenuse is 6 cm long, and ∠A measures 32°. Find the length of the shorter leg. 58°

6

Make a sketch 32° ■ If one angle is 32°, the other is 58° ■ The shorter leg is opposite the smaller angle, so you need to find the side opposite the 32° angle. ■

©Carolyn C. Wheater, 2000

9

Choosing the Ratio ◆ ... Find the length of the shorter leg. You need a ratio that talks about opposite and hypotenuse ■ Can use sine (sin) or cosecant (csc), but since your calculator has a key for sin, sine is more convenient. ■

©Carolyn C. Wheater, 2000

58°

6 32°

10

Solving the Triangle x sin( 32 ) = 6 58°

6 32°

From your calculator, you can find that sin(32°) ≈ 0.53, so x 0.53 = 6

x ≈ 3.2 ©Carolyn C. Wheater, 2000

11

Special Right Triangles ◆ 45 – 45 – 90 Triangle The legs are of equal length ■ The length of the hypotenuse is 2 ■

s

s s

2

45 °

s

2 sin( 45 ) = = 2 s 2 s

2 cos( 45 ) = = 2 s 2

s tan( 45 ) = = 1 s

times the leg ©Carolyn C. Wheater, 2000

12

Special Right Triangles ◆ 30 – 60 – 90 Triangle The side opposite the 30° angle is half the hypotenuse ■ The side opposite the 60° angle is half the hypotenuse times 3 ■

1 2

30 °

h

1 sin( 30 ) = = h 2 1

h 3 3 cos( 30 ) = = h 2 2

1 2

h

3 tan( 30 ) = 1 = 3 h 3 2

©Carolyn C. Wheater, 2000

13

60 °

Special Right Triangles ◆ 30 – 60 – 90 Triangle The side opposite the 30° angle is half the hypotenuse ■ The side opposite the 60° angle is half the hypotenuse times 3

1

sin( 60 ) =

2

h 3 h

■

1 2

3 = 2

h

1 cos( 60 ) = = h 2 1

tan( 60 ) =

©Carolyn C. Wheater, 2000

2

h 3 1 2

h

= 3

14

Memory Work ◆ Know these values as well as you know your own name. sin

cos

tan

30°

1 2

3 2

3 3

45°

2 2

2 2

1

60°

3 2

1 2

©Carolyn C. Wheater, 2000

3

15

Non-Right Triangles ◆All these relationships are based on the assumption that the triangle is a right triangle. ◆It is possible, however, to use trigonometry to solve for unknown sides or angles in non-right triangles. ©Carolyn C. Wheater, 2000

16

Law of Sines

a b c = = sin( A) sin( B) sin( C)

◆ In geometry, you learned that the largest angle of a triangle was opposite the longest side, and the smallest angle opposite the shortest side. ◆ The Law of Sines says that the ratio of a side to the sine of the opposite angle is constant throughout the triangle. ©Carolyn C. Wheater, 2000

17

Sample Problem ◆In ∆ABC, m∠A = 38°, m∠B = 42°, and BC = 12 cm. Find the length of side AC. ■ Draw

a diagram to see the position of the C given angles and side. ■ BC is opposite ∠A A ■ You must find AC, the side opposite ∠B. ©Carolyn C. Wheater, 2000

18

B

Sample Problem ◆.... Find the length of side AC. ■ Use

the Law of Sines with m∠A = 38°, m∠B = 42°, and BC = 12 a b = sin( A) sin( B)

12 b = sin( 38 ) sin( 42 )

12 sin( 42 ) = b sin( 38 )

b g

12 sin( 42 ) 12 .6691 b= ≈ sin( 38 ) .6157 b ≈ 13.041

©Carolyn C. Wheater, 2000

19

Warning ◆The Law of Sines is useful when you know ■ the

sizes of two sides and one angle or ■ two angles and one side.

◆However, the results can be ambiguous if the given information is two sides and an angle other than the included angle (ssa). ©Carolyn C. Wheater, 2000

20

Warning ◆The Law of Sines gives a unique solution when the given information is ■ sas ■ asa

Remember that these are all sufficient conditions for congruent triangles.

■ aas

◆The ambiguous case is ssa, which is not a way of proving triangles congruent. ©Carolyn C. Wheater, 2000

21

Law of Cosines c 2 = a 2 + b 2 − 2ab cos(C )

◆ If you apply the Law of Cosines to a right triangle, that extra term becomes zero, leaving just the Pythagorean Theorem. ◆ The Law of Cosines is most useful when you know the lengths of all three sides and need to find an angle, or ■ when you two sides and the included angle. ■

©Carolyn C. Wheater, 2000

22

Largest angle opposite longest side

Sample Problem

◆Triangle XYZ has sides of lengths 15, 22, and 35. Find the measure of the largest angle of the triangle. c 2 = a 2 + b 2 − 2ab cos( C) 22

15 35

352 = 152 + 22 2 − 2 ⋅15 ⋅ 22 ⋅ cos( C ) 1225 = 225 + 484 − 660 cos( C) 1225 = 709 − 660 cos( C) ©Carolyn C. Wheater, 2000

23

Sample Problem ◆... Find the measure of the largest angle of the triangle. 516= − 660 cos( C) 22

15 35

−

516 − cos( C) = = .7818 660 C = cos−1 ( − .7818) ≈ 1414 . ©Carolyn C. Wheater, 2000

24

Caution ◆Many people, when they reach the line 1225 = 709 − 660 cos(C ) will mistakenly subtract 660 from 709.

◆Don’t be one of them. ◆The multiplication should be done before any addition or subtraction. ©Carolyn C. Wheater, 2000

25