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The Law of Cosines
The Law of Cosines
If A, B, mid C are the measures of the angles of a
triangle, and a, b, and c are the lengths of the sides
opposite these angles, then
a2 = b2 + c2 − 2bc cos A
b2 = a2 + c2 − 2ac cos B
c2 = a2 + b2 − 2ab cos C.
The square of a side of a triangle equals the sum of
the squares of the other two sides minus twice
their product times the cosine of their included
angle.
Solving an SAS Triangle
1. Use the Law of Cosines to find the side opposite
the given angle.
2. Use the Law of Sines to find the angle opposite
the shorter of the two given sides. This angle is
always acute.
3. Find the third angle. Subtract the measure of the
given angle and the angle found in step 2 from
180º.
1
Text Example
C
b = 20
• Solve the triangle shown with A = 60º, b = 20, and c = 30.A
a
60º
c =30
B
Solution
We are given two sides and an included angle. Therefore, we
apply the three−step procedure for solving a SAS triangle.
Step l Use the Law of Cosines to find the side opposite the given angle.
Thus, we will find a.
a2 = b2 + c2 − 2bc cos A
Apply the Law of
Cosines to find a.
b = 20, c = 30, and A = 60°.
a2 = 202 + 302 − 2(20)(30) cos 60º
= 400 + 900 − 1200(0.5) = 700
Perform the indicated operations.
a =  700 = 26
Take the square root of both sides
and solve for a.
Text Example cont.
C
b = 20
Solve the triangle shown with A = 60º, b = 20, and c = 30.
A
a
60º
c=
30
B
Solution
Step 2 Use the Law of Sines to find the angle opposite the shorter
of the two given sides. This angle is always acute. The shorter of the
two given sides is b = 20. Thus, we will find acute angle B.
b
a
=
sin B sin A
700
20
=
sin B sin 60o
700 sin B = 20sin 60o
sin B =
B
41
šo
B Šš
41
20sin60 o
≈ 0.6547
700
Apply the Law of Sines.
We are given b = 20 and A = 60°. Use
the exact value of a,  700, from step 1.
Cross multiply.
Divide by square root of 700 and solve
for sin B.
Find sin-10.6547 using a calculator.
Text Example cont.
C
Solve the triangle shown with A = 60º, b = 20, and c = 30.
b = 20
A
a
60º
c=
30
B
Solution
We are given two sides and an included angle. Therefore, we
apply the three−step procedure for solving a SAS triangle.
Step 3 Find the third angle. Subtract the measure of the given angle and
the angle found in step 2 from 180º.
C = 180º − A − B = 180º − 60º − 41º = 79º
The solution is a = 26, B = 41º, and C = 79º.
2
Solving an SSS Triangle
1. Use the Law of Cosines to find the angle
opposite the longest side.
2. Use the Law of Sines to find either of the
two remaining acute angles.
3. Find the third angle. Subtract the measures
of the angles found in steps 1 and 2 from
180º.
Text Example
• Two airplanes leave an airport at the same time on
different runways. One flies at a bearing of N66ºW at 325
miles per hour. The other airplane flies at a bearing of
S26ºW at 300 miles per hour. How far apart will the
airplanes be after two hours?
Solution
After two hours. the plane flying at 325
miles per hour travels 325 · 2 miles, or 650 miles.
Similarly, the plane flying at 300 miles per hour
travels 600 miles. The situation is illustrated in the
figure.
Let b = the distance between the planes after two
hours. We can use a north−south line to find angle B
in triangle ABC. Thus,
B = 180º − 66º − 26º = 88º.
We now have a = 650, c = 600, and B = 88º.
Text Example cont.
Two airplanes leave an airport at the same time on different
runways. One flies at a bearing of N66ºW at 325 miles per
hour. The other airplane flies at a bearing of S26ºW at 300
miles per hour. How far apart will the airplanes be after two
hours?
Solution
We use the Law of Cosines to find b in this SAS situation.
b2 = a2 + c2 − 2ac cos B
Apply the Law of Cosines.
b2 = 6502 + 6002 − 2(650)(600) cos 88º
= 755,278
b = 869
Substitute: a= 650, c =600, and B= 88°.
Use a calculator.
Take the square root and solve for b.
After two hours, the planes are approximately 869 miles apart.
3
Heron’s Formula
The area of a triangle with sides a, b, and c is
Area = s ( s − a )( s − b)( s − c)
s=
1
(a + b + c)
2
Example
• Use Heron’s formula to find the area of the
given triangle:
a=10m, b=8m, c=4m
Solution:
1
(a + b + c)
2
1
s = (10 + 8 + 4)
2
1
s = ( 22) = 11
2
s=
Area = s ( s − a )( s − b)( s − c)
= 11(11 − 10)(11 − 8)(11 − 4)
= 11(1)(3)(7) = 231 sq.m.
The Law of Cosines
4