# QF Q2.) Solve the triangle

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AsdfjkadfjkakjfC=180-A-B=180-67-53=60deg. by sine rule b/sinB=a/sinA b=a/sinA*sinB =13/sin(67)*sin(53) = 11.28 c/sinC=a/sinA c=a/sinA*sinC =13/sin(67)*sin(60) = 12.23

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nustopocruSum of interior angles in a triangle is 180 deg.so angle C = 180 - A - B=180 - 67 - 53 =60deg. Use the sine rule in a triangle - sinA / a = sinB / b b = a * sinB / sinA =13 * sin(53) / sin(67) =11.3 Similarly sinC / c = sinA / a c = a * sinC / sinA =13 * sin(60) / sin(67) =12.2