Question 546343: A triangular lot has sides of 200 meters, 180 meters, and 120 meters. Find the measure of the largest angle. I tried solving this problem by using the Law of Cosine- a^2=b^2+c^2-2(b)(c)cosA : 200^2=180^2+120^2-2(180)(120)cosA: 40,000=32,400+14,400-43,200cosA: 40,000=46,800-43,200cosA: 40,000-46,800=-43,200cosA: -6800=-43,200cosA: -6800/-43,200=cosA: 17/108=cosA That's all I could do. I don' know how to find the cosine of a letter.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A triangular lot has sides of 200 meters, 180 meters, and 120 meters.
Find the measure of the largest angle. I tried solving this problem by using the Law of Cosine-
a^2=b^2+c^2-2(b)(c)cosA :
200^2=180^2+120^2-2(180)(120)cosA:
40,000=32,400+14,400-43,200cosA:
40,000=46,800-43,200cosA:
40,000-46,800=-43,200cosA:
-6800=-43,200cosA:
-6800/-43,200=cosA:
17/108=cosA
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A = cos^-1(17/108) = 80.94 degrees
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Note: cos(A) = (b^2+c^2-a^2)/(2bc)
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cos(A) = (180^2+120^2-200^2)/(2*180*120) = 0.1574
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A = cos^-1(0.1574) = 80.94 degrees
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Cheers,
Stan H.
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