McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. The Law of Sines and Law of Cosines
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Exercise 33 Page 676

Begin by using the Law of Cosines.

m ∠ J ≈ 65
m ∠ K ≈ 66
m ∠ L ≈ 49

Practice makes perfect

Let's consider the given diagram. We will use the same color for a side and its opposite vertex. This will help us use the Law of Cosines and later the Law of Sines.

First, we can tell that this is not a right triangle, as the sides do not satisfy the Pythagorean Theorem. 24.6^2+ 29.7^2 ≠ 30.0^2 Let's find the measures of ∠ J, ∠ K, and ∠ L one at a time.

Finding m ∠ J

The lengths of all three sides of the triangle are given. Therefore, we can use the Law of Cosines to find m ∠ J. j^2=k^2+l^2 -2 k l cos J Let's substitute j= 29.7, k= 30.0, and l= 24.6 to isolate cos J.
j^2=k^2+l^2 -2 k l cos J
29.7^2= 30.0^2+ 24.6^2 -2 ( 30.0)( 24.6) cos J
Solve for cos J
882.09=900+605.16-2(30.0)(24.6)cos J
882.09 = 900 + 605.16 -1476 cos J
882.09=1505.16-1476 cos J
1476 cos J +882.09=1505.16
1476 cos J=623.07
cos J =623.07/1476
Now, we can use the inverse cosine ratio and a calculator to find m ∠ J.
m ∠ J = cos ^(-1) (623.07/1476)
m ∠ J = 65.030601...
m ∠ J ≈ 65

Finding m ∠ K

Now that we know the measure of ∠ J, we can find m ∠ K using the Law of Sines. sin J/j =sin K/k Let's substitute j= 29.7, m ∠ J ≈ 65, and k= 30, to isolate sin K.
sin J/j =sin K/k
sin 65/29.7 = sin K/30
Solve for sin K
sin 65/29.7* 30 = sin K
30 sin 65/29.7 = sin K
sin K =30 sin 65/29.7
Now we can use the inverse sine ratio to find m ∠ K.
m ∠ K = sin ^(-1) (30 sin 65/29.7)
m ∠ K = 66.271481...
m ∠ K ≈ 66

Finding m ∠ L

Finally, to find m ∠ L we can use the Triangle Angle Sum Theorem. This tells us that the measures of the angles in a triangle add up to 180. 65+ 66 + m ∠ L = 180 ⇔ m ∠ L ≈ 49

Completing the Triangle

With all of the angle measures, we can complete our diagram.