McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
5. Law of Cosines
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Exercise 18 Page 826

Begin by using the Law of Sines if two angles and any side are given, or if two sides and an angle opposite one of them are given. Begin by using the Law of Cosines if two sides and their included angle are given, or if the three sides are given.

Law: Cosines
Missing Parts: s≈ 28.9, R≈ 42^(∘), T≈ 32^(∘)

Practice makes perfect

We can use the Law of Sines and the Law of Cosines to find missing measurements in triangles that are not right triangles. Finding all of the missing measurements is called solving the triangle. To do this, we have to determine which of the two laws must be used first.

Solving Triangles
Given Begin by Using
Two angles and any sides Law of Sines
Two sides and an angle opposite one of them Law of Sines
Two sides and their included angle Law of Cosines
Three sides Law of Cosines

Let's determine which law to use first and then solve the triangle.

Determining the Law

In △ R S T we are told that S= 106^(∘), r= 20, and t= 16.

We have information about two sides and their included angle. Therefore, we must start solving the triangle by using the Law of Cosines.

Solving the Triangle

We will find the missing side length and the missing angle measures one at a time.

Finding the Side Length

Let's consider the Law of Cosines and adapt it to our triangle. s^2= r^2+ t^2-2 r tcos S We can substitute the given values into this equation and solve for s.

s^2=r^2+t^2-2rtcos S
s^2= 20^2+ 16^2-2( 20)( 16)cos 106^(∘)
â–¼
Solve for s
s^2 = 400 + 256 - 640cos 106^(∘)
s^2 = 400 + 256 - 640(- 0.275637...)
s^2 = 400 + 256 + 176.407907...
s^2 = 832.407907...
s = sqrt(832.407907...)
s = 28.851480...
s ≈ 28.9

Finding the Angle Measures

Let's add the newly obtained information to our diagram.

Since now we have three side lengths and one angle measure, we can use the Law of Sines. sin R/r=sin S/s Let's substitute r= 20, S= 106^(∘), and s= 28.9 into the above equation and solve for R.

sin R/r=sin S/s
sin R/20=sin 106^(∘)/28.9
â–¼
Solve for R
sin R=sin 106^(∘)/28.9 * 20

sin ^(- 1)(LHS)=sin ^(- 1)(RHS)

R = sin ^(- 1) ( sin 106^(∘)/28.9 * 20 )
R = 41.700203... ^(∘)
R ≈ 42^(∘)

Finally, we can use the Triangle Angle Sum Theorem to find the measure of T. 42^(∘) + 106^(∘) + T = 180^(∘) ⇕ T= 32^(∘) We have completely solved the triangle.