Question

Solve △ABC if B=120°, a=10, c=18

Answer 1

`b= √(a^2+c^2-2ac*cosB) = √(10^2+18^2-2*10*18*(-0.5)) = \\ = √(100+324+180)= √(604) \approx 24.58 \ units \\ \\ \\ (b)/(sinB) =(a)/(sinA) \ \ \to sinA= (a*sinB)/(b)= (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= (1)/(2)ch \\ \\ h=a*sinB=10*sin120^o=10* 0.866=8.66 \ units \\ \\ Area= (1)/(2)*18*8.66=77.94 \ units^2`