# Thread: Prove the sine rule

1. ## Prove the sine rule

Prove the sine rule for a triangle ABC by drawing AD perpendicular to BC meeting BC at D. Consider two cases (a) B acute (b) B obtuse.

I managed to prove the $\displaystyle \frac{b}{\sin B}=\frac{c}{\sin C}$
but i can't prove that that is equal to $\displaystyle \frac{a}{\sin A}$
i cant do part (b).
$\displaystyle \sin B=\frac{AD}{AB}$ and $\displaystyle \sin C=\frac{AD}{AC}$
but i can't find such an equation for A.
Thanks for any help.

2. Hello arze
Originally Posted by arze
Prove the sine rule for a triangle ABC by drawing AD perpendicular to BC meeting BC at D. Consider two cases (a) B acute (b) B obtuse.

I managed to prove the $\displaystyle \frac{b}{\sin B}=\frac{c}{\sin C}$
but i can't prove that that is equal to $\displaystyle \frac{a}{\sin A}$
i cant do part (b).
$\displaystyle \sin B=\frac{AD}{AB}$ and $\displaystyle \sin C=\frac{AD}{AC}$
but i can't find such an equation for A.
Thanks for any help.
The case where $\displaystyle B$ is obtuse simply gives:
$\displaystyle AD = c\sin(180^o-B) = b\sin C$
and then using $\displaystyle \sin(180^o-B) = \sin B$, the result follows immediately.

I'm not sure what is expected if you are to show that each of these is equal to $\displaystyle \frac{a}{\sin A}$. I should be quite happy to accept the argument:
Similarly, by dropping the perpendicular from $\displaystyle B$ to $\displaystyle AC$: $\displaystyle \frac{c}{\sin C}=\frac{a}{\sin A}$
This is much the most straightforward method. However, if you are not allowed to do that, and we are given the formula
$\displaystyle \sin(x+y)=\sin x \cos y + \cos x \sin y$
then the only thing I can suggest is that we denote $\displaystyle \angle BAD=A_B$ and $\displaystyle \angle CAD=A_C$, and then say:
$\displaystyle \frac{a}{\sin A}=\frac{a}{\sin(A_B+A_C)}$
$\displaystyle =\frac{a}{\sin A_B \cos A_C + \cos A_B \sin A_C}$

$\displaystyle =\frac{a}{\dfrac{BD}{c}\dfrac{AD}{b}+\dfrac{AD}{c} \dfrac{DC}{b}}$

$\displaystyle =\frac{abc}{AD(BD+DC)}$

$\displaystyle =\frac{abc}{c\sin B(a)}$

$\displaystyle =\frac{b}{\sin B}$
But that seems excessively complicated!

3. Originally Posted by arze
Prove the sine rule for a triangle ABC by drawing AD perpendicular to BC meeting BC at D. Consider two cases (a) B acute (b) B obtuse.

I managed to prove the $\displaystyle \frac{b}{\sin B}=\frac{c}{\sin C}$
but i can't prove that that is equal to $\displaystyle \frac{a}{\sin A}$
i cant do part (b).
$\displaystyle \sin B=\frac{AD}{AB}$ and $\displaystyle \sin C=\frac{AD}{AC}$
but i can't find such an equation for A.
Thanks for any help.

Hi arze,

you can also use the attached sketches to guide you,
rewriting the angles, vertices and sides for your own notation.

Once we have $\displaystyle \frac{SinA}{a}=\frac{SinB}{b}$ and $\displaystyle \frac{SinC}{c}=\frac{SinB}{b}$

then since $\displaystyle \frac{SinB}{b}$ is equal to both $\displaystyle \frac{SinA}{a}$ and $\displaystyle \frac{SinC}{c}$ then they are also equal, so $\displaystyle \frac{SinA}{a}=\frac{SinB}{b}=\frac{SinC}{c}$

In the obtuse case, you only need use the trigonometric identity $\displaystyle Sin(180^o-B)=SinB$ or $\displaystyle Sin(A-B)=SinACosB-CosASinB$

For $\displaystyle A=180^o$, this is $\displaystyle Sin(180)CosB-Cos(180)SinB=0--SinB=SinB$