13-6The
13-6
TheLaw
LawofofCosines
Cosines
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Algebra
Holt
Algebra
22
13-6 The Law of Cosines
Warm Up
Find each measure to the nearest tenth.
1. my 104°
2. x ≈ 8.8
3. y ≈ 18.3
4. What is the area of ∆XYZ ? Round to the
nearest square unit. 60 square units
Holt Algebra 2
13-6 The Law of Cosines
Objectives
Use the Law of Cosines to find the side
lengths and angle measures of a
triangle.
Use Heron’s Formula to find the area
of a triangle.
Holt Algebra 2
13-6 The Law of Cosines
In the previous lesson, you learned to solve
triangles by using the Law of Sines. However, the
Law of Sines cannot be used to solve triangles for
which side-angle-side (SAS) or side-side-side (SSS)
information is given. Instead, you must use the
Law of Cosines.
Holt Algebra 2
13-6 The Law of Cosines
To derive the Law of Cosines, draw ∆ABC with
altitude BD. If x represents the length of AD, the
length of DC is b – x.
Holt Algebra 2
13-6 The Law of Cosines
Write an equation that relates the side lengths
of ∆DBC.
a2 = (b – x)2 + h2
Pythagorean Theorem
a2 = b2 – 2bx + x2 + h2
Expand (b – x)2.
2 = x2 + h2.
In
∆ABD,
c
=
– 2bx +
Substitute c2 for x2 + h2.
a2 = b2 – 2b(c cos A) + c2 In ∆ABD, cos A = or x =
cos A. Substitute
2
2
2
a = b + c – 2bccos A
c cos A for x.
a2
b2
c2
The previous equation is one of the formulas for the
Law of Cosines.
Holt Algebra 2
13-6 The Law of Cosines
Holt Algebra 2
13-6 The Law of Cosines
Example 1A: Using the Law of Cosines
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
a = 8, b = 5, mC = 32.2°
Step 1 Find the length of the third side.
c2 = a2 + b2 – 2ab cos C
Law of Cosines
c2 = 82 + 52 – 2(8)(5) cos 32.2°
Substitute.
c2 ≈ 21.3
c ≈ 4.6
Holt Algebra 2
Use a calculator to
simplify.
Solve for the positive
value of c.
13-6 The Law of Cosines
Example 1A Continued
Step 2 Find the measure of the smaller angle, B.
Law of Sines
Substitute.
Solve for sin B.
Solve for m
Holt Algebra 2
B.
13-6 The Law of Cosines
Example 1A Continued
Step 3 Find the third angle measure.
mA + 35.4° + 32.2°  180°
mA  112.4°
Holt Algebra 2
Triangle Sum Theorem
Solve for m A.
13-6 The Law of Cosines
Example 1B: Using the Law of Cosines
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
a = 8, b = 9, c = 7
Step 1 Find the measure of the largest angle,
B.
b2 = a2 + c2 – 2ac cos B
Law of cosines
92 = 82 + 72 – 2(8)(7) cos B
cos B = 0.2857
Substitute.
Solve for cos B.
m
Solve for m B.
Holt Algebra 2
B = Cos-1 (0.2857) ≈ 73.4°
13-6 The Law of Cosines
Example 1B Continued
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
Step 2 Find another angle measure
c2 = a2 + b2 – 2ab cos C
Law of cosines
72 = 82 + 92 – 2(8)(9) cos C
Substitute.
cos C = 0.6667
Solve for cos C.
m
Solve for m C.
Holt Algebra 2
C = Cos-1 (0.6667) ≈ 48.2°
13-6 The Law of Cosines
Example 1B Continued
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
Step 3 Find the third angle measure.
m
A + 73.4° + 48.2°  180°
m
Holt Algebra 2
A  58.4°
Triangle Sum Theorem
Solve for m A.
13-6 The Law of Cosines
Check It Out! Example 1a
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
b = 23, c = 18, m
A = 173°
Step 1 Find the length of the third side.
a2 = b2 + c2 – 2bc cos A
Law of Cosines
a2 = 232 + 182 – 2(23)(18) cos 173° Substitute.
a2 ≈ 1672.8
a ≈ 40.9
Holt Algebra 2
Use a calculator to
simplify.
Solve for the positive
value of c.
13-6 The Law of Cosines
Check It Out! Example 1a Continued
Step 2 Find the measure of the smaller angle, C.
Law of Sines
Substitute.
Solve for sin C.
m
C = Sin-1
Holt Algebra 2
Solve for m
C.
13-6 The Law of Cosines
Check It Out! Example 1a Continued
Step 3 Find the third angle measure.
m
B + 3.1° + 173°  180°
m
Holt Algebra 2
B  3.9°
Triangle Sum Theorem
Solve for m B.
13-6 The Law of Cosines
Check It Out! Example 1b
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
a = 35, b = 42, c = 50.3
Step 1 Find the measure of the largest angle,
c2 = a2 + b2 – 2ab cos C
C.
Law of cosines
50.32 = 352 + 422 – 2(35)(50.3) cos C Substitute.
cos C = 0.1560
Solve for cos C.
m
Holt Algebra 2
C = Cos-1 (0.1560) ≈ 81.0°
Solve for m C.
13-6 The Law of Cosines
Check It Out! Example 1b Continued
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
a = 35, b = 42, c = 50.3
Step 2 Find another angle measure
a2 = c2 + b2 – 2cb cos A
352 = 50.32 + 422 – 2(50.3)(42) cos A
Law of cosines
Substitute.
cos A = 0.7264
Solve for cos A.
m
Solve for m A.
A = Cos-1 (0.7264) ≈ 43.4°
Holt Algebra 2
13-6 The Law of Cosines
Check It Out! Example 1b Continued
Step 3 Find the third angle measure.
m
B + 81° + 43.4°  180°
m
Holt Algebra 2
B  55.6°
Solve for m B.
13-6 The Law of Cosines
Remember!
The largest angle of a triangle is the angle
opposite the longest side.
Holt Algebra 2
13-6 The Law of Cosines
Example 2: Problem-Solving Application
If a hiker travels at an
average speed of 2.5 mi/h,
how long will it take him to
travel from the cave to the
waterfall? Round to the
nearest tenth of an hour.
1
Understand the Problem
The answer will be the number of hours that
the hiker takes to travel to the waterfall.
Holt Algebra 2
13-6 The Law of Cosines
1
Understand the Problem
The answer will be the number of hours that
the hiker takes to travel to the waterfall.
List the important information:
• The cave is 3 mi from the cabin.
• The waterfall is 4 mi from the cabin. The path
from the cabin to the waterfall makes a 71.7°
angle with the path from the cabin to the cave.
• The hiker travels at an average speed of 2.5 mi/h.
Holt Algebra 2
13-6 The Law of Cosines
2
Make a Plan
Use the Law of Cosines to find the distance d
between the water-fall and the cave. Then
determine how long it will take the hiker to
travel this distance.
Holt Algebra 2
13-6 The Law of Cosines
3
Solve
Step 1 Find the distance d between the waterfall
and the cave.
d2 = c2 + w2 – 2cw cos D
d2 = 42 + 32 – 2(4)(3)cos 71.7°
d2 ≈ 17.5
d ≈ 4.2
Holt Algebra 2
Law of Cosines
Substitute 4 for c,
3 for w, and
71.7 for D.
Use a calculator to simplify.
Solve for the positive
value of d.
13-6 The Law of Cosines
Step 2 Determine the number of hours.
The hiker must travel about 4.2 mi to
reach the waterfall. At a speed of 2.5
mi/h, it will take the hiker
≈ 1.7 h
to reach the waterfall.
4
Look Back
In 1.7 h, the hiker would travel 1.7  2.5
= 4.25 mi. Using the Law of Cosines to
solve for the angle gives 73.1°. Since this
is close to the actual value, an answer of
1.7 hours seems reasonable.
Holt Algebra 2
13-6 The Law of Cosines
Check It Out! Example 2
A pilot is flying from
Houston to Oklahoma City.
To avoid a thunderstorm,
the pilot flies 28° off the
direct route for a distance
of 175 miles. He then
makes a turn and flies
straight on to Oklahoma
City. To the nearest mile,
how much farther than the
direct route was the route
taken by the pilot?
Holt Algebra 2
13-6 The Law of Cosines
1
Understand the Problem
The answer will be the additional distance
the pilot had to fly to reach Oklahoma City.
List the important information:
• The direct route is 396 miles.
• The pilot flew for 175 miles off the course
at an angle of 28° before turning towards
Oklahoma City.
Holt Algebra 2
13-6 The Law of Cosines
2
Make a Plan
Use the Law of Cosines to find the distance
from the turning point on to Oklahoma City.
Then determine the difference additional
distance and the direct route.
Holt Algebra 2
13-6 The Law of Cosines
3
Solve
Step 1 Find the distance between the turning
point and Oklahoma City. Use side-angle-side.
b2 = c2 + a2 – 2ca cos B
Law of Cosines
b2 = 3962 + 1752 –
2(396)(175)cos 28°
Substitute 396 for c, 175
for a, and 28° for B.
b2 ≈ 65072
b ≈ 255
Holt Algebra 2
Use a calculator
to simplify.
Solve for the positive
value of b.
13-6 The Law of Cosines
Step 2 Determine the number of additional
miles the plane will fly.
Add the actual miles flown and subtract
from that normal distance to find the extra
miles flown
255 + 175 = 430
Total miles traveled.
430 – 396 = 34
Additional miles.
4
Look Back
By using the Law of Cosines the length of the
extra leg of the trip could be determined.
Holt Algebra 2
13-6 The Law of Cosines
The Law of Cosines can be used to derive a formula
for the area of a triangle based on its side lengths.
This formula is called Heron’s Formula.
Holt Algebra 2
13-6 The Law of Cosines
Example 3: Landscaping Application
A garden has a triangular flower bed with
sides measuring 2 yd, 6 yd, and 7 yd. What is
the area of the flower bed to the nearest
tenth of a square yard?
Step 1 Find the value of s.
Use the formula for half of
the perimeter.
Substitute 2 for a, 6 for b,
and 7 for c.
Holt Algebra 2
13-6 The Law of Cosines
Example 3 Continued
Step 2 Find the area of the triangle.
A=
Heron’s formula
A=
Substitute 7.5
for s.
Use a calculator
to simplify.
A = 5.6
The area of the flower bed is 5.6 yd2.
Holt Algebra 2
13-6 The Law of Cosines
Example 3 Continued
Check Find the measure of the largest angle, C.
c2 = a2 + b2 – 2ab cos C
Law of Cosines
72 = 22 + 62 – 2(2)(6) cos C
Substitute.
Solve for cos C.
cos C ≈ –0.375
m C ≈ 112.0°
Solve for m
Find the area of the triangle by using
the formula area = ab sin c.
area
Holt Algebra 2

c.
13-6 The Law of Cosines
Check It Out! Example 3
The surface of a hotel swimming pool is shaped
like a triangle with sides measuring 50 m, 28 m,
and 30 m. What is the area of the pool’s surface
to the nearest square meter?
Step 1 Find the value of s.
Use the formula for half of
the perimeter.
Substitute 50 for a, 28 for
b, and 30 for c.
Holt Algebra 2
13-6 The Law of Cosines
Check It Out! Example 3 Continued
Step 2 Find the area of the triangle.
A=
A = 367 m2
Heron’s formula
Substitute 54 for
s.
Use a calculator
to simplify.
The area of the flower bed is 367 m2.
Holt Algebra 2
13-6 The Law of Cosines
Check It Out! Example 3 Continued
Check Find the measure of the largest angle, A.
502 = c2 + b2 – 2cb cos A
Law of Cosines
502 = 302 + 282 – 2(30)(28) cos A
Substitute.
Solve for cos A.
cos A ≈ –0.4857
m A ≈ 119.0°
Solve for m
Find the area of the triangle by using
the formula area = ab sin A.

Holt Algebra 2
A.
13-6 The Law of Cosines
Lesson Quiz: Part I
Use the given measurements to solve ∆ABC.
Round to the nearest tenth.
1. a = 18, b = 40, m C = 82.5°
c ≈ 41.7; m A ≈ 25.4°; m B ≈ 72.1°
2. a = 18.0; b = 10; c = 9
m A ≈ 142.6°; m B ≈ 19.7°; m
Holt Algebra 2
C ≈ 17.7°
13-6 The Law of Cosines
Lesson Quiz: Part II
3. Two model planes take off from the same spot.
The first plane travels 300 ft due west before
landing and the second plane travels 170 ft
southeast before landing. To the nearest foot,
how far apart are the planes when they land?
437 ft
4. An artist needs to know the area of a
triangular piece of stained glass with sides
measuring 9 cm, 7 cm, and 5 cm. What is the
area to the nearest square centimeter?
17 cm2
Holt Algebra 2