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

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Copyright © 2011 Pearson, Inc. 5.6 Law of Cosines

Transcript of 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.

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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

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Example Solving a Triangle (SAS)

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

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Example Solving a Triangle (SAS)

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

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

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Example Solving a Triangle (SAS)

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

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

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Example Solving a Triangle (SSS)

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

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Example Solving a Triangle (SSS)

Use Law of Cosines to find one angle:

a2 =b2 + c2 −2bccosA

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

A=cos−17380

⎛⎝⎜

⎞⎠⎟≈24.1º

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

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Example Solving a Triangle (SSS)

Use Law of Cosines to find a second angle:

a2 =b2 + c2 −2bccosA

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

B=cos−1 −0.19( ) ≈101.0º

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

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Example Solving a Triangle (SSS)

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

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

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Area of a Triangle

ΔArea=

12

bcsinA=12

acsinB=12

absinC

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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( ).

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Example Using Heron’s Formula

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

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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.

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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.

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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⎛

⎝⎜⎞

⎠⎟

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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

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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

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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.

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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?

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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.

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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

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Chapter Test Solutions

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

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Chapter Test Solutions

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

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Chapter Test Solutions

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