# Angles in triangles are in AP, find:$\dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C$

If angles of a $\triangle{ABC}$ are in A.P. then find the value of : $$\dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C$$ where $a,b,c$ are sides and $A,B,C$ are angles.

My attempts:

Let $\angle A =x-d , \angle B=x, \angle C=x+d\ \implies \angle B=60^{o}$.

So one possibility is following:

1. $\angle A=30^{o}$
2. $\angle C=90^{o}$

By sine law: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$

$\implies \dfrac{a}{c}\sin 2A+\dfrac{c}{a}\sin 2C=\dfrac{\sin^{2}A \sin 2A+ \sin^{2} C \sin 2C}{\sin A \sin C}$ By value of $\angle A\ ,\angle C$.

Answer comes to be $\dfrac{\sqrt{3}}{4}$.

Is it correct, if not then where I go wrong? And I don't want to prove this by an assumption that $\angle A=30, \angle C=90$. How to do this without assuming angles.

Edit as suggested by @rohan :

original question is follows:

If angles of a $\triangle{ABC}$ are in A.P. then find the value of : $$\dfrac{a}{c}\sin 2C+\dfrac{c}{a}\sin 2A$$ where $a,b,c$ are sides and $A,B,C$ are angles.

I earnestly apologise for wasting your precious time, it was not deliberate.

• Another case is $A=B=C=60^o$. – Nosrati Dec 31 '16 at 8:14
• Yes, it is a 2010 problem. Please edit the question. – Rohan Dec 31 '16 at 8:29
• @Rohan ok thanks. – mnulb Dec 31 '16 at 8:31

I think there is a typo in your question. It should be to find the value of $\frac{a}{c}\sin 2C + \frac{c}{a}\sin 2A$.
Assuming the case it is so, then the problem simplifies to: $$\frac{\sin A}{\sin C}(2\sin C\cos C) + \frac{\sin C}{\sin A}(2\sin A\cos A) = 2\sin A\cos C + 2\sin C\cos A = 2\sin (A+C) = 2\sin 2B = 2\times \frac{\sqrt{3}}{2} =\sqrt{3}$$
• I think there is a minor typo, $A+C=2B$ – Siong Thye Goh Dec 31 '16 at 8:21