Question

In a triangle ABC, prove that sin A/sin(A+B)=a/c​

Answer 1

Answer:

Step-by-step explanation:

a/sinA=b/sinB=c/sinC=2R

so, a=2RsinA

c=2RsinC

solving L.H.S.

»a/c

»2RsinA/2RsinC

»sinA/sinC

»sinA/sin(180-C) (because sin(180-x)=sinx)

»sinA/sin(A+B)=RHS

because A+B+C=180

SO 180-C=A+B

Answer 2

$\dfrac{\sin A}{\sin(A+B)}=\dfrac{a}{c}$, proved.

Step-by-step explanation:

To prove that $\dfrac{\sin A}{\sin(A+B)}=\dfrac{a}{c}$.

R.H.S. = $\dfrac{a}{c}$

Using sine rule,

$\dfrac{a}{\sin A} =\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=k$

∴ $a = k\sin A, b = k\sin B and c = k\sin C$

$= \dfrac{a}{c}$

$= \dfrac{k\sin A}{k\sin C}$

∵ A + B + C = 180°

⇒ C = 180° - (A + B)

$= \dfrac{\sin A}{\sin (180° - (A + B))}$

Using the trigonometric identity,

$\sin (180-\theta)=\sin \theta$

$= \dfrac{\sin A}{\sin(A+B)}$

= L.H.S., proved.

Thus, $\dfrac{\sin A}{\sin(A+B)}=\dfrac{a}{c}$, proved.