Question

Solve △ABC if B=120°, a=10, c=18

Answer 1

$b= \sqrt{a^2+c^2-2ac*cosB} = \sqrt{10^2+18^2-2*10*18*(-0.5)} = \\ = \sqrt{100+324+180}= \sqrt{604} \approx 24.58 \ units \\ \\ \\ \frac{b}{sinB} =\frac{a}{sinA} \ \ \to sinA= \frac{a*sinB}{b}= \frac{10*0.866}{24.58} \approx 0.3523 \ \to \ \angle{A} \approx 20.63^o \\ \\ \\ \angle{C}=180-120-20.63=39.37^o \\ \\ \\ Area= \frac{1}{2}ch \\ \\ h=a*sinB=10*sin120^o=10* 0.866=8.66 \ units \\ \\ Area= \frac{1}{2}*18*8.66=77.94 \ units^2$