9.6 apply the law of cosines in which cases can the law of cosines be used to solve a triangle? what...

9.6 Apply the Law of Cosines

In which cases can the law of cosines be used to solve a triangle?

What is Heron’s Area Formula?What is the semiperimeter?

Law of Cosines

Use the law of cosines to solve triangles when two sides and the included angle are known (SAS), or when all three side are known (SSS).

Solve ABC with a = 11, c = 14, and B = 34°.

SOLUTION

Use the law of cosines to find side length b.

b2 = a2 + c2 – 2ac cos B

b2 = 112 + 142 – 2(11)(14) cos 34°

b2 61.7

b2 61.7 7.85

Law of cosines

Substitute for a, c, and B.

Simplify.

Take positive square root.

Use the law of sines to find the measure of angle A.

sin Aa

sin Bb

=

sin A11

=sin 34°7.85

sin A =11 sin 34°

7.850.7836

A sin –1 0.7836 51.6°

Law of sines

Substitute for a, b, and B.

Multiply each side by 11 andSimplify.

Use inverse sine.

The third angle C of the triangle is C 180° – 34° – 51.6° = 94.4°.

In ABC, b 7.85, A 51.68, and C 94.48.ANSWER

Solve ABC with a = 12, b = 27, and c = 20.

SOLUTION

First find the angle opposite the longest side, AC . Use the law of cosines to solve for B.

b2 = a2 + c2 – 2ac cos B

272 = 122 + 202 – 2(12)(20) cos B

272 = 122 + 202

– 2(12)(20)= cos B

– 0.3854 cos BB cos –1 (– 0.3854) 112.7°

Law of cosines

Substitute.

Solve for cos B.

Simplify.

Use inverse cosine.

Now use the law of sines to find A.

sin Aa =

sin Bb

sin A12

sin 112.7°27

=

sin A =12 sin 112.7°

270.4100

A sin–1 0.4100 24.2°

Law of sines

Substitute for a, b, and B.

Multiply each side by 12 and simplify.

Use inverse sine.

The third angle C of the triangle is C 180° – 24.2° – 112.7° = 43.1°.

In ABC, A 24.2, B 112.7, and C 43.1.ANSWER

ScienceScientists can use a set of footprints to calculate an organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to 180°, the more efficiently the organism walked.

The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B.

SOLUTION b2 = a2 + c2 – 2ac cos B3162 = 1552 + 1972 – 2(155)(197) cos B

3162 = 1552 + 1972

– 2(155)(197)= cos B

– 0.6062 cos B

B cos –1 (– 0.6062) 127.3° Use inverse cosine.

Simplify.

Solve for cos B.

Substitute.

Law of cosines

The step angle B is about 127.3°.ANSWER

Find the area of ABC.

1. a = 8, c = 10, B = 48°

SOLUTION

Use the law of cosines to find side length b.

b2 = a2 + c2 – 2ac cos B

b2 = 82 + 102 – 2(8)(10) cos 48°

b2 57

b2 57 7.55

Law of cosines

Substitute for a, c, and B.

Simplify.

Take positive square root.

Use the law of sines to find the measure of angle A.

sin Aa

sin Bb

=

sin A 8

=sin 48°7.55

sin A =8 sin 48°

7.550.7874

A sin –1 0.7836 51.6°

Law of sines

Substitute for a, b, and B.

Multiply each side by 8 andsimplify.

Use inverse sine.

The third angle C of the triangle is C 180° – 48° – 51.6° = 80.4°.

In ABC, b 7.55, A 51.6°, and C 80.4°.ANSWER

162 = 142 + 92 – 2(14)(9) cos B

Find the area of ABC. 2. a = 14, b = 16, c = 9

SOLUTION First find the angle opposite the longest side, AC . Use the law of cosines to solve for B.

b2 = a2 + c2 – 2ac cos B

162 = 142 + 92

– 2(14)(9)= cos B

Law of cosines

Substitute.

Solve for cos B.

– 0.0834 cos BB cos –1 (– 0.0834) 85.7°

Simplify.

Use inverse cosine.

sin Aa

= sin Bb

sin A14

sin 85.2°16

=

sin A =14sin 85.2°

160.8719

Law of sines

Substitute for a, b, and B.

Multiply each side by 14 and simplify.

Use the law of sines to find the measure of angle A.

The third angle C of the triangle is C 180° – 85.2° – 60.7° = 34.1°.

A sin–1 0.8719 60.7° Use inverse sine.

In ABC, A 60.7°, B 85.2°, and C 34.1°.

ANSWER

Heron’s Area FormulaHeron (Hero) of Alexandria, the Greek mathematician (10 - 70 A.D.) is credited with using the law of cosines to find this formula for the area of a triangle.

Urban PlanningThe intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown.

SOLUTION STEP 1 Find the semiperimeter s.

s = (a + b + c )12

12= (170 + 240 + 350) = 380

STEP 2 Use Heron’s formula to find the area of ABC.

Area = s (s – a) (s – b) (s – c)

380 (380 – 170) (380 – 240) (380 – 350)= 18,300

The area of the traffic triangle is about 18,300 square yards.

Find the area of ABC. 4.

SOLUTION

STEP 1 Find the semiperimeter s.

s = (a + b + c )12

12= (5 + 8 + 11) = 12

STEP 2 Use Heron’s formula to find the area of ABC.

Area = s (s – a) (s – b) (s – c)

The area is about 18.3 square units.

12 (12 – 8) (12 – 11) (12 – 5)= 18.3

Find the area of ABC. 5.

SOLUTION

STEP 1 Find the semiperimeter s.

s = (a + b + c )12

12= (4 + 9+ 7) = 10

STEP 2 Use Heron’s formula to find the area of ABC.

Area = s (s – a) (s – b) (s – c)

The area is about 13.4 square units.

10 (10– 4) (10 – 9) (10 – 7)= 13.4

Find the area of ABC.6.

SOLUTION

STEP 1 Find the semiperimeter s.

s = (a + b + c )12

12= (15 + 23+ 127) = 25

STEP 2 Use Heron’s formula to find the area of ABC.

Area = s (s – a) (s – b) (s – c)

The area is about 80.6 square units.

25 (25– 15) (25 – 23) (25 – 12)= 80.6

9.6 Assignment

Page 596, 3-39 every 3rd problem No work is the same as a missing problem