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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.
Slide 5.6 - 3 Copyright © 2011 Pearson, Inc.
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
Slide 5.6 - 4 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle (SAS)
Solve ΔABC given that a=10, b=4 and ∠C =25o.
Slide 5.6 - 5 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 6 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 7 Copyright © 2011 Pearson, Inc.
Example Solving a Triangle (SSS)
Solve ΔABC given that a=5, b=12 and c=10.
Slide 5.6 - 8 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 9 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 10 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 11 Copyright © 2011 Pearson, Inc.
Area of a Triangle
ΔArea=
12
bcsinA=12
acsinB=12
absinC
Slide 5.6 - 12 Copyright © 2011 Pearson, Inc.
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( ).
Slide 5.6 - 13 Copyright © 2011 Pearson, Inc.
Example Using Heron’s Formula
Find the area of a triangle with sides 10, 12, 14.
Slide 5.6 - 14 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 15 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 16 Copyright © 2011 Pearson, Inc.
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⎛
⎝⎜⎞
⎠⎟
Slide 5.6 - 17 Copyright © 2011 Pearson, Inc.
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
Slide 5.6 - 18 Copyright © 2011 Pearson, Inc.
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
Slide 5.6 - 19 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 20 Copyright © 2011 Pearson, Inc.
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?
Slide 5.6 - 21 Copyright © 2011 Pearson, Inc.
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.
Slide 5.6 - 22 Copyright © 2011 Pearson, Inc.
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
Slide 5.6 - 23 Copyright © 2011 Pearson, Inc.
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
Slide 5.6 - 24 Copyright © 2011 Pearson, Inc.
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