Question

Hitunglah besar sudut segitiga ABC ,jika diketahui A=5x, B=3x+50°,dan C=2x+20°

Answer 1

<A+<B+<C = 180
5x+3x+50+2x+20 = 180
10x = 180-70
10x = 110
x = 11

<A = 5.11 = 55⁰
<B = 3.11+50 = 33+50 = 83⁰
<C = 2.11+20 = 22+20 = 42⁰

Answer 2

A + b + c = 180
5x + 3x +50 + 2x+20 = 180
10x = 180-50-20
10x = 110
x = 110/10
x = 11

a = 5x = 5 . 11 = 55
b = 3x + 50 = 3 . 11 + 50 = 33 + 50 = 83
c = 2x + 20 = 2 . 11 + 20 = 22 + 20 = 42