How do you show that for triangle ABC, $\frac{a}{c} = \frac{sin (B + C)}{sin C}$ $a/c=(sin(B+C))/sinC$ ?

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Answer 1 · 2016-10-18T15:15:40

See below.

For any triangle is true the sinus law

$\frac{a}{sin (A)} = \frac{b}{sin (B)} = \frac{c}{sin (C)}$ $a/sin(A)=b/sin(B)=c/sin(C)$

but also

$\frac{a}{c} = \frac{sin (A)}{sin (C)}$ $a/c=sin(A)/sin(C)$

We know also that $A + B + C = π$ $A+B+C=pi$

so

$\frac{a}{c} = \frac{sin (π - (A + B))}{sin (c)} = \frac{sin (A + B)}{sin (C)}$ $a/c=sin(pi-(A+B))/sin(c)=sin(A+B)/sin(C)$

How do you show that for triangle ABC, ac=sin(B+C)sinCa/c=(sin(B+C))/sinC?