Question

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

Answer 1

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

Answer 2

$\frac{\sin A}{\sin (A+B)}$ = $\frac{a}{c}$

Step-by-step explanation:

Given Data

ToProve  - $\frac{\sin A}{\sin (A+B)}$ =   $\frac{a}{c}$ for a triangle ABC

The Formula for Alternative sine rule is

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2 R$

Where, $\frac{a}{\sin A}$ = 2R

then a = 2R sinA -----------> (1)

similarly, $\frac{C}{\sin C}$ = 2R

c = 2R sinC -------------> (2)

on dividing (1) by (2)

$\frac{a}{c}=\frac{2 \mathrm{R} \sin A}{2 R \sin C}$  -----------> (3)

We know that the trigonometric formula for sin (180 - theta) is equal to sin theta

then sin (180 - C ) = sin C

substitute sin C = sin (180 - C ) in (3)

$\frac{a}{c}=\frac{\sin A}{\sin (180-C)}$  ------------> (4)

For a triangle , A + B + C = 180

Also, A + B = 180 - C

substitute 180 - C = A + B in  (4)

$\frac{a}{c}=\frac{\sin A}{\sin (A+B)}$

Hence Proved that  $\frac{a}{c}=\frac{\sin A}{\sin (A+B)}$

To learn more ...

1. https://brainly.in/question/169507

2. https://brainly.in/question/1171496