Copyright © 2011 Pearson, Inc. 5.6 Law of Cosines.

Copyright © 2011 Pearson, Inc.

5.6Law of Cosines

Slide 5.6 - 2 Copyright © 2011 Pearson, Inc.

What you’ll learn about

Deriving the Law of Cosines Solving Triangles (SAS, SSS) Triangle Area and Heron’s Formula Applications

… and whyThe Law of Cosines is an important extension of the Pythagorean theorem, with many applications.


Law of Cosines

Let ΔABC be any triangle with sides and angles labeled in the usual way. Then

a2 =b2 + c2 −2bccosA

b2 =a2 + c2 −2accosB

c2 =a2 +b2 −2abcosC


Example Solving a Triangle (SAS)

Solve ΔABC given that a=10, b=4 and ∠C =25o.



Use the Law of Cosines to find side c:

c2 =a2 +b2 −2abcosC

c2 =16 +100−2(4)(10)cos25o

c=6.6

Use the Law of Cosines again:

102 =16 + 43.56 −2(4)(6.6)cosAcosA=−0.7659

A=140o




Now find (sum of the angles in a triangle = 180º):

∠B=180o−140o−25o =15o

The six parts of the triangle are:

∠A=140o a=10,

∠B=15o b=4,

∠C =25o c=6.6



Example Solving a Triangle (SSS)

Solve ΔABC given that a=5, b=12 and c=10.



Use Law of Cosines to find one angle:


52 =122 +102 −2 12( ) 10( )cosA240cosA=219

A=cos−17380

⎛⎝⎜

⎞⎠⎟≈24.1º




Use Law of Cosines to find a second angle:


122 =52 +102 −2 5( ) 10( )cosB100cosB=−19

B=cos−1 −0.19( ) ≈101.0º




Now find (sum of the angles in a triangle = 180º):

∠C =180o−140o−25o =15o

The six parts of the triangle are:

∠A≈14.1o a=5,

∠B≈110.0o b=12,

∠C ≈54.9o c=10



Area of a Triangle

ΔArea=

12

bcsinA=12

acsinB=12

absinC


Heron’s Formula

Let a, b, and c be the sides of ΔABC, and let s denote

the semiperimeter (a+b+ c) / 2. Then the area of

ΔABC is given by

Area= s s−a( ) s−b( ) s−c( ).


Example Using Heron’s Formula

Find the area of a triangle with sides 10, 12, 14.


Example Using Heron’s Formula

Find the area of a triangle with sides 10, 12, 14.

Compute s: s =(10 +12 +14) / 2 =18.Use Heron's Formula:

A= 18 18−10( ) 18−12( ) 18−14( )

= 3456

=24 6 ≈58.8The area is approximately 58.8 square units.


Quick Review

Find an angle between 0o and 180o that is a solution

to the equation.

1. cos A =4 / 52. cos A=-0.25Solve the equation (in terms of x and y) for

(a) cos A and (b) A, 0 ≤A≤180o.

3. 72 =x2 + y2 −2xycosA

4. y2 =x2 + 4−4xcosA5. Find a quadratic polynomial with real coefficients

that has no real zeros.


Quick Review

Find an angle between 0o and 180o that is a solution

to the equation.

1. cos A =4 / 5 36.87o

2. cos A=−0.25 104.48o

Solve the equation (in terms of x and y) for

(a) cos A and (b) A, 0 ≤A≤180o.

3. 72 =x2 + y2 −2xycosA

(a) 49−x2 −y2

−2xy (b) cos-1

49−x2 −y2

−2xy⎛

⎝⎜⎞

⎠⎟


Quick Review

Solve the equation (in terms of x and y) for

(a) cos A and (b) A, 0 ≤A≤180o.

4. y2 =x2 + 4−4xcosA

(a) y2 −x2 −4

−4x (b) cos-1

y2 −x2 −4−4x

⎛

⎝⎜⎞

⎠⎟

5. Find a quadratic polynomial with real coefficients

that has no real zeros.

One answer: x2 + 2


Chapter Test

1. Prove the identity cos3x =4cos3 x−3cosx.

2. Write the expression in terms of sinx and cosx.

cos2 2x−sin2x

3. Find the general solution without using a calculator.2cos2x=1


Chapter Test

4. Solve the equation graphically. Find all solutions

in the interval [0,2π ). sin4 x+ x2 =25. Find all solutions in the interval [0,2π ) without

using a calculator. sin2 x−2sinx−3=06. Solve the inequality. Use any method, but give

exact answers. 2cosx<1 for 0 ≤x< 2π7. Solve ΔABC, given A=79o, B=33o, and a=7.8. Find the area of ΔABC, given a=3, b=5, and c=6.


Chapter Test

9. A hot-air balloon is seen over Tucson, Arizona, simultaneously by two observers at points A and B that are 1.75 mi apart on level ground and in line with the balloon. The angles of elevation are as shown here. How high above ground is the balloon?


Chapter Test

10. A wheel of cheese in the shape of a right circular cylinder is 18 cm in diameter and 5 cm thick. If a wedge of cheese with a central angle of 15º is cut from the wheel, find the volume of the cheese wedge.


Chapter Test Solutions

1. Prove the identity cos3x =4cos3 x−3cosx.cos3x=cos(2x+ x) =cos2xcosx−sin2xsinx

= cos2 x−sin2 x( ) cosx( )− 2sinxcosx( )sinx

=cos3 x−3cosxsin2 x

=cos3 x−3cosx 1−cos2 x( ) =4cos3 x−3cosx.

2. Write the expression in terms of sinx and cosx.

cos2 2x−sin2x 1−4sin2 xcos2 x−2cosxsinx3. Find the general solution without using a calculator.

2cos2x=1 π6+ nπ, 5π

6+ nπ



4. Solve the equation graphically. Find all solutions

in the interval [0,2π ). sin4 x+ x2 =2 x≈1.155. Find all solutions in the interval [0,2π ) without

using a calculator. sin2 x−2sinx−3=0 3π2

6. Solve the inequality. Use any method, but give

exact answers. 2cosx<1 for 0 ≤x< 2π π3,5π3

⎛

⎝⎜⎞

⎠⎟

7. Solve ΔABC, given A=79o, B=33o, and a=7.

C ≈68o, b≈3.88, c≈6.61



9. A hot-air balloon is seen over Tucson, Arizona, simultaneously by two observers at points A and B that are 1.75 mi apart on level ground and in line with the balloon. The angles of elevation are as shown here. How high above ground is the balloon?

≈ 0.6 mi

8. Find the area of VABC, given a =3, b=5, and c=6. ≈7.5



10. A wheel of cheese in the shape of a right circular cylinder is 18 cm in diameter and 5 cm thick. If a wedge of cheese with a central angle of 15º is cut from the wheel, find the volume of the cheese wedge.

405π/24 ≈ 53.01

Copyright © 2011 Pearson, Inc. 5.6 Law of Cosines.

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