The triangle PQR isn't a square triangle. PQ = 9.9cm QR = 22.3cm RP = 18.1cm (Going around the triangle anti clockwise) P Q R Calculate the area of the triangle. Please help me and state how you did it
Follow Math Help Forum on Facebook and Google+
Cosine rule.... a^2 = b^2 + c^2 - 2bcCosA 2bcCosA = b^2 + c^2 - a^2 CosA = (b^2 + c^2 - a^2) / 2bc CosA = (9.9^2 + 22.3^2 - 18.1^2) / 441.51 A = arccos((9.9^2 + 22.3^2 - 18.1^2) / 441.51) = 0.919389657 radians Area = 0.5bcSinA = 87.8 cm^2
Originally Posted by iwz The triangle PQR isn't a square triangle. PQ = 9.9cm QR = 22.3cm RP = 18.1cm (Going around the triangle anti clockwise) P Q R Calculate the area of the triangle. Please help me and state how you did it This problem can also be solved using Heron's formula and a,b,c are the side lengths of the triangle. so the Area is
Originally Posted by sean.1986 Cosine rule.... a^2 = b^2 + c^2 - 2bcCosA 2bcCosA = b^2 + c^2 - a^2 CosA = (b^2 + c^2 - a^2) / 2bc CosA = (9.9^2 + 22.3^2 - 18.1^2) / 441.51 A = arccos((9.9^2 + 22.3^2 - 18.1^2) / 441.51) = 0.919389657 radians Area = 0.5bcSinA = 87.8 cm^2 Thanks, that's what i was looking for
View Tag Cloud