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!

Grandad