Given that Triangle ABC such that

a=8cm, b=7cm and c=9cm

tem and c = 9cm find Cos B

We use cosine rule also called cosine law which states

c^2=a^2+b^2-2abcosC

given

a=8cm, b=7cm, c=9cm cos C?

9^2=8^+7^2-2*8*7 cos C

expand

81=64+49-112 cos C

like terms together

81-113=-112 cos C

-32=-112cos C

multiply both sides by - 1

32=112cos C

divide both sides by 112

32/112=cos C

cosC=0.2857

find cos inverse of 0.2857

angle C = 73.40