Section 6.2
The Law of Cosines
About the Law of Cosines
• Can be used when given two sides and the
angle between them (SAS) or when given
three sides (SSS) but no angles.
• No ambiguous case.
• Make sure your calculator is in DEGREE mode.
• Once you identify a matched pair (an angle
and its opposite side), you can resume using
the Law of Sines.
LAW OF COSINES
Law of Cosines : a 2  b 2  c 2 - 2bc cos A
b 2  a 2  c 2 - 2ac cos B
c 2  a 2  b 2 - 2ab cos C
SAS
SSS
If the known parts of the triangle are SAS OR
SSS, use the Law of Cosines. The known parts
of shown in red in the triangles above.
Example 1: Solve the triangle where a = 6, b = 9, and c = 4
(Give angle measures to the nearest degree.)
This information gives SSS. We will need to use the Law of
Cosines to find an angle. Find the largest angle first. The
largest angle is opposite the longest side.
b2 = a2 + c2 -2ac cosB
92 = 62 + 42 -2(6)(4) cosB
b2 = a2 + c2 -2ac cosB
92  6  42
cos B 
2(6)(4)
B = 127⁰. (continued)
 92  6  42 
B  cos 


2(6)(4)


1
Example 1 continued
You may use the Law of Cosine or the Law of Sine to find
another angle. The Law of Sine requires less work!
6
9

sin A sin127
C ≈ 180⁰ - 127⁰ - 32⁰ ≈ 21⁰
A ≈ 320
Example 2
Solve the triangle with A = 60 degrees, b = 20, c = 30.
Round sides to the nearest tenth and angles to the nearest degree.
Example 3
Two ships leave a harbor at the same time. One ship
travels on a bearing of S12⁰W at 14 miles per hours. The
other ship travels on a bearing of N75⁰E at 10 miles per
hour. How far apart will the ships be after three hours:
(Round to the nearest tenth of a mile.)
Finding Area of an Oblique Triangle when the 3 sides are known.
Heron’s Formula:
1
Area  s( s  a)(s  b)(s  c), where s   a  b  c 
2
Example 4:
Find the area of a triangle having side lengths of a = 6 meters, b = 16
meters, and c = 18 meters. Round to the nearest square meter.
Find s first. It is half of the perimeter of the triangle.
s = (1/2)(6 + 16 + 18) = 20. Now use s in Heron’s Formula.
Area  20(20  6)(20  16)(20 18)
Area ≈ 47 m2
Example 5
A piece of commercial real estate is priced at $3.50 per
square foot. Find the cost, to the nearest dollar, of a
triangular lot measuring 240 feet by 300 feet by 420
feet.
Example 6
A baseball diamond has four bases forming a square whose sides
measure 90 feet each. The pitcher’s mound is 60.5 feet from home
plate on a line joining home plate and second base. Find the
distance from the pitcher’s mound to first base. Round to the
nearest tenth of a foot.