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

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Answer 1 · 2016-10-18T15:15:40

See below.

For any triangle is true the sinus law

$a/sin(A)=b/sin(B)=c/sin(C)$

but also

$a/c=sin(A)/sin(C)$

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

so

$a/c=sin(pi-(A+B))/sin(c)=sin(A+B)/sin(C)$

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