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 27 Page 826

Use the Law of Cosines to find one of the angle measures in the triangle. Then use the formula for the area of a triangle using sine.

13 148 yd^2

Practice makes perfect

We are told that some land is in the shape of a triangle and we are given the distances between each vertex of this triangle. We want to find the area of the land using the Law of Cosines. Let's start by making a diagram.

To find the area of the triangle we will use the formula for the area of a triangle using sine.

Before using this formula, we need to know one of the angle measures of our triangle. This is where we will use the Law of Cosines. Since angle ∠ B is the largest, we will find its measure. b^2= a^2+ c^2-2 a ccos B Let's substitute a= 210, b= 300, and c= 140 to isolate cos B.

b^2=a^2+c^2 -2 a c cos B
300^2= 210^2+ 140^2 -2 ( 210)( 140) cos B
â–¼
Solve for cos B
90 000=44 100+19 600-2(210)(140)cos B
90 000=44 100+19 600-58 800cos B
90 000=63 700-58 800cos B
26 300=- 58 800cos B
26 300/- 58 800=cos B
-26 300/58 800=cos B
- 0.447278...=cos B
cos B=0.447278...
cos B ≈ - 0.4473

Now, we can use the inverse cosine ratio and a calculator to find m ∠ B.

m ∠ B = cos ^(-1) (- 0.4473)
m ∠ B ≈ 116.570586...
m ∠ B ≈ 116.57

Now, knowing the measure of ∠ B we can use the formula for the area of a triangle using sine with the sine of ∠ B. Area=1/2 a csin B Let's substitute a= 210, c= 140 and m∠ B= 116.57^(∘).

Area=1/2ac sin B
Area=1/2( 210)( 140) sin 116.57^(∘)
â–¼
Simplify right-hand side
Area=1/2(29 400) sin 116.57^(∘)
Area=29 400/2 sin 116.57^(∘)
Area=14 700 sin 116.57^(∘)
Area=14 700(0.894388... )
Area=13 147.5036...
Area≈ 13 148

The area of the land is approximately 13 148 square yards.