Question

HELP PLEASE..answer fast please and show all work and explain how to solve

Answer 1

`\cfrac{a}{sinA}= \cfrac{b}{sinB} \ \ \to \\ sinB= \cfrac{b*sinA}{a} = \cfrac{10*sin15^o}{9}= 0.2875 \ \ \to \ m \angle B=16.7^o \\ \\ \\ m \angle C=180- m \angle A- m \angle B=180-15-16.7=148.3^o \\ \\ \\ \cfrac{a}{sinA}= \cfrac{c}{sinC} \ \ \to \ c= \cfrac{a*sinC}{sinA}= \cfrac{9*sin148.3^o}{sin15^o} \approx 18.3`