<s>[INST] Given the following relation between the angles of a triangle, find the number of possible triangles. If $A,B,C$ are the interior angles of triangle $\Delta ABC$ such that
$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=9$, then the number of possible triangles is
$(A)0\hspace{1cm}(B)1\hspace{1cm}(C)3\hspace{1cm}(D)\infty$

$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B+2\cos B\cos C+2\cos C\cos A+\sin^2A+\sin^2B+\sin^2C+2\sin A\sin B+2\sin B\sin C+2\sin C\sin A=9$
I do not understand how I can identify the number of possible triangles from this tedious expression.
 [/INST] $$(\cos A+\cos B+\cos C)^2+(\sin A+\sin B+\sin C)^2=9$$
$$\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B+2\cos B\cos C+2\cos C\cos A+\sin^2A+\sin^2B+\sin^2C+2\sin A\sin B+2\sin B\sin C+2\sin C\sin A=9$$
$$3+2(\cos A\cos B+\cos B\cos C+\cos C\cos A+\sin A\sin B+\sin B\sin C+\sin C\sin A)=9$$
$$\cos A\cos B+\cos B\cos C+\cos C\cos A+\sin A\sin B+\sin B\sin C+\sin C\sin A=3$$
$$\cos(A-B)+\cos(C-B)+\cos(A-C)=3$$
Then
$A+B+C=\pi$ and $A=B=C \Rightarrow A=B=C=60^{\circ}$
Answer: $(B)$ $1$
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