MR-697 / 16.) problem

Two sides of a triangle are 10 cm and 14 cm, and the angle opposite the first is 45°. Calculate the area of the triangle.

sin 45°=hcb

hc=b·sin 45°

hc=14·22=7·2·22

hc=72

a2=b2+c2-2bc·cos 45°

102=142+c2-2·14·c·22

100=196+c2-142c

196+c2-142c-100=0

c2-142c+96=0

c1,2=--142±1422-4·1·962

c1,2=142±196·2-3842

c1,2=142±82

c1,2=142±2·42=142±242

c1,2=142±222

c1=142+222=1622=8·222

c1=82

c2=142-222=1222=6·222

c2=62

A1=c1·hc2=82·722=8·7

A1=56

A2=c2·hc2=62·722=6·7

A2=42