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 36 Page 827

Think about a right triangle where you are given all of the side lengths.

See solution.

Practice makes perfect

Let's imagine that one the rooms of our house is in the shape of a right triangle. We want to find its angle measures to design a special carpet for the room.

With all of the side lengths, we can find the corresponding angles one at a time. To do so, we will start by recalling the Law of Cosines. a^2= b^2+ c^2-2 b c * cos A b^2= a^2+ c^2-2 a c * cos B c^2= a^2+ b^2-2 a c * cos CLet's use the first formula to find the value of m ∠ A. We will substitute a= 4, b= 5, and c= 3 into the formula.

a^2=b^2+c^2-2bc * cos A
4^2= 5^2+ 3^2-2( 5) ( 3) * cos A
â–¼
Solve for A
16=25+9-2(5)(3) * cos A
16=25+9-30 * cos A
16=34-30 * cos A
16+ 30 * cos A =34
30 * cos A=18
cos A=18/30
cos A=3/5

Then we need to use inverse cosine to isolate the measure of the angle. cos A=3/5 ⇔ m ∠ A=cos^(-1)(3/5) To do so we will use a calculator. We need to first push MODE in our calculator and change Radian to Degree.

Window with a graph

Then, we push 2ND followed by COS, introduce the value 35, and push ENTER.

The inverse cosine of 35 is about 53^(∘), which corresponds to the value of m ∠ A. With this value we can find the third angle measure B. Let's use the Triangle Angle Sum Theorem. 53 ^(∘) + 90 ^(∘) + m ∠ C = 180 ^(∘) ⇓ m ∠ C ≈ 37 ^(∘) Great! Now, we can complete our right triangle.