# SOLUTION: 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=

Algebra ->  Algebra  -> Customizable Word Problem Solvers  -> Geometry -> SOLUTION: 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=      Log On

 Ad: Over 600 Algebra Word Problems at edhelper.com Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Word Problems: Geometry Solvers Lessons Answers archive Quiz In Depth

 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(57387)   (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 --- A = cos^-1(17/108) = 80.94 degrees ======================================= Note: cos(A) = (b^2+c^2-a^2)/(2bc) --- cos(A) = (180^2+120^2-200^2)/(2*180*120) = 0.1574 --- A = cos^-1(0.1574) = 80.94 degrees ========================================== Cheers, Stan H. -----