If A, B, C are interior angles of △ABC such that (cos A -cosB + cosC)2 +(sin A + sinB + sinC)2 =9, then number of possible triangles is

(Σcos⁡A)2+(Σsin⁡A)2=9Σcos2⁡A+sin2⁡A+2(Σcos⁡Acos⁡B+sin⁡Asin⁡B)3+2Σcos⁡(A−B)≤3+2(3)=9Equality holds if A = B = C⇒ △ABC is equilateral ⇒ Infinite many equilateral[Note We can vary side length of equilateral triangle]

If A, B, C are interior angles of △ABC such that (cos A -cosB + cosC)2 +(sin A + sinB + sinC)2 =9, then number of possible triangles is

(Σcos⁡A)2+(Σsin⁡A)2=9Σcos2⁡A+sin2⁡A+2(Σcos⁡Acos⁡B+sin⁡Asin⁡B)3+2Σcos⁡(A−B)≤3+2(3)=9Equality holds if A = B = C⇒ △ABC is equilateral ⇒ Infinite many equilateral[Note We can vary side length of equilateral triangle]

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If A, B, C are interior angles of $△$ ABC such that (cos A -cosB + cosC)² +(sin A + sinB + sinC)² =9, then number of possible triangles is

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infinite

answer is D.

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Detailed Solution

$(Σcos A)^{2} + (Σsin A)^{2} = 9$
$Σ (\cos^{2} A + \sin^{2} A) + 2 (Σcos Acos B + \sin Asin B) 3 + 2 Σcos (A - B) \leq 3 + 2 (3) = 9$
Equality holds if A = B = C
$\Rightarrow$ $△$ ABC is equilateral $\Rightarrow$ Infinite many equilateral
[Note We can vary side length of equilateral triangle]