Question

In a triangle ABC if A,B,C ar in A.P with sin(2A+B)=sin(C-A)=sin(B+2C)=1/2 then what are the values of A,B and C?

Answer 1

In a triangle ABC if A,B,C are in A.P with $\sin(2A+B)=\sin(C-A)=\sin(B+2C)=1/2$ then what are the values of A,B and C?

Solution:

If A, B, C are in A.P then

$$A = a,$$ $$B = a + b,$$ $$C = a + 2b.$$

Suppose that $A \leq B \leq C$. Then $b \geq 0$. Also we have

$$A + B + C = \pi,$$ $$a + a + b + a + 2b = \pi,$$ $$3a + 3b = \pi,$$ $$a + b = \frac{\pi}{3}.$$

So

$$B = a + b = \frac{\pi}{3}$$

Further

$$\sin(2A + B) = \frac{1}{2}.$$

Because $A > 0$ and $B = \frac{\pi}{3}$ then

$$2A + \frac{\pi}{3} = \pi - \frac{\pi}{6},$$ $$A = \frac{\pi}{4}$$

So

$$\left\{ \begin{array}{c} a = \frac{\pi}{4}, \\ a + b = \frac{\pi}{3}. \end{array} \right.$$

Thus we have $b = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$ and

$$C = a + 2b = \frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}.$$

By the condition of the task we have

$$\sin (B + 2 C) = \frac {1}{2}.$$

But there is

$$\sin (B + 2 C) = \sin \left(\frac {\pi}{3} + \frac {5 \pi}{6}\right) = - \frac {1}{2}.$$

Thus values of A, B, C do not exist.

**Answer**: Values of A, B, C do not exist.