An equation with angles of triangle satisfies 60 degrees triad, can we say that the triangle is equilateral? Question:
In a triangle ABC, the equation $2\cos A \sin C=\sin B $ holds true. What type of triangle is it?
Choices available are: a) isosceles b) equilateral
Solution:
$$2\cos A \sin C=\sin B $$
$$\Rightarrow \sin(A+C) -\sin (A-C)=\sin B$$
$$\Rightarrow \sin(A+C) -\sin (A-C)=\sin(180º - (A+C))$$
$$\Rightarrow \sin(A+C) -\sin (A-C)=\sin (A+C)$$
$$\Rightarrow \sin (A-C),  A = C$$
Hence isosceles.
Confusion:
But, using $A=B=C=\frac{\pi}{3}$ in the original equation, we get $2*\frac{1}{2}*\frac{\sqrt3}{2}=\frac{\sqrt3}{2}$ which is true.
I tried to obtain a proof for both &quot;the triangle is equilateral&quot; and &quot;the triangle is not equilateral&quot;, but I end in deadlock situation, so could not obtain any proof for any of the two claims.
So, just because the equation satisfies, $A=B=C=\frac{\pi}{3}$, can we say that the triangle is equilateral?
Textbook has &quot;isosceles&quot; as the answer.
You have successfully proved that $A=C$. However, you cannot show that $A=B=C$. So the triangle cannot be shown to be equilateral. However, every equilateral triangle, also has both sides equal, so equilateral triangles also satisfy your condition. So what we have is,
Every equilateral triangle satisfies your equation, but not every triangle that satisfies your equation is equilateral, but it is definitely isosceles. So the correct answer would be isosceles.
Your solution is right.
Your example is also right because an equilateral triangle is also an isosceles triangle.
Formally you can say that the triangle isosceles or equilateral, as for example, you can say that the triangle is isosceles or right-angled isosceles triangle with $\beta=90^{\circ},$ but it's better to say that the answer it's: The triangle is an isosceles triangle.