law of cosines - HCC Learning Web

3.2
LAW OF COSINES
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Introduction
Two cases remain in the list of conditions needed to solve
an oblique triangleβ€”SSS and SAS.
If you are given three sides (SSS), or two sides and their
included angle (SAS), none of the ratios in the Law of
Sines would be complete.
Introduction
In such cases, you can use the Law of Cosines.
Example Three Sides of a Triangleβ€”SSS
Find the three angles of the triangle.
It is a good idea first to find the angle opposite the longest
sideβ€”side b in this case. Using the alternative form of the
Law of Cosines, you find that
Example
cos 𝐡 β‰ˆ βˆ’0.45089
Because cos B is negative, you know that B is an obtuse
angle given by B ο‚» 116.80ο‚°.
At this point, it is simpler to use the Law of Sines to
determine A.
Example
sin 𝐴 β‰ˆ 0.37583
You know that A must be acute because B is obtuse, and
a triangle can have, at most, one obtuse angle.
So, A ο‚» 22.08ο‚° and
C ο‚» 180ο‚° – 22.08ο‚° – 116.80ο‚°
= 41.12ο‚°
Example
SAS
Use the Law of Cosines to solve the triangle with the
following measurements: 𝐴 = 48°, 𝑏 = 3, 𝑐 = 14
Using the Law of Cosines we can find the missing side, a
π‘Ž2 = 𝑏 2 + 𝑐 2 βˆ’ 2𝑏𝑐 cos 𝐴
π‘Ž2 = 32 + 142 βˆ’ 2(3)(14)(cos 48°)
= 9 + 196 βˆ’ 84(.6691)
= 205 βˆ’ 56.207
= 148.79
π‘Ž = 12.20
Take square root of both sides
Example SAS
Now we have a side and opposite angle
Use Law of Sines
sin 𝐴 sin 𝐡
=
π‘Ž
𝑏
β†’
sin 48° sin 𝐡
=
12.2
3
sin 𝐡 =
sin 48°
12.2
3
sin 𝐡 = 0.182741
𝐡 = 10.53°
Example
Lastly, for angle C
𝐢 = 180° βˆ’ 𝐴 βˆ’ 𝐡
𝐢 = 180° βˆ’ 48° βˆ’ 10.53°
𝐢 = 121.47°
Heron’s Area Formula
Heron’s Area Formula
The Law of Cosines can be used to establish the following
formula for the area of a triangle.
This formula is called Heron’s Area Formula
Example 5 – Using Heron’s Area Formula
Find the area of a triangle having sides of lengths
a = 43 meters, b = 53 meters, and c = 72 meters.
Because s = (a + b + c)/2
= 168/2
= 84,
Heron’s Area Formula yields
Example 5 – Solution
π΄π‘Ÿπ‘’π‘Ž =
𝑠 𝑠 βˆ’ π‘Ž 𝑠 βˆ’ 𝑏 (𝑠 βˆ’ 𝑐)
π΄π‘Ÿπ‘’π‘Ž =
84(84 βˆ’ 43)(84 βˆ’ 53)(84 βˆ’ 72)
π΄π‘Ÿπ‘’π‘Ž =
84(41)(31)(12)
Area β‰ˆ 1131.89 π‘šπ‘’π‘‘π‘’π‘Ÿπ‘  2
cont’d
Area Formulae
You have now studied three different formulas for the area
of a triangle.
Standard Formula:
Oblique Triangle:
1
2
Area = (π‘π‘Žπ‘ π‘’)(β„Žπ‘’π‘–π‘”β„Žπ‘‘)
Area =
1
𝑏𝑐 sin 𝐴
2
1
2
= π‘Žπ‘ sin 𝐢
1
2
= π‘Žπ‘ sin 𝐡
Heron’s Area Formula: Area =
𝑠(𝑠 βˆ’ π‘Ž)(𝑠 βˆ’ 𝑏)(𝑠 βˆ’ 𝑐)
In Class
Section 3.2, Pg 295, 296 # 5, 13, 33, 43