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QJEEAdv23Mab3a2: JEE Advanced 2023 Mathematics - Class 11 JEEAdv23Mab3a2 Solution: Consider an obtuse-angled triangle A B C in which the…

Consider an obtuse-angled triangle ABC in which the difference between the largest and the smallest angle is π2 and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.  Let a be the area of the triangle ABC. Then the value of (64a)2 is:

Step-by-Step Verified Solution

Area of △ABC=AB⋅BC⋅AC4R\triangle A B C=\frac{A B \cdot B C \cdot A C}{4 R}

⇒a=8sin⁡A⋅sin⁡Bsin⁡C4=2sin⁡(π2−2C)sin⁡(π2+C)sin⁡C=2cos⁡2C⋅cos⁡C⋅sin⁡C=cos⁡2C⋅sin⁡2C=1−sin2⁡2C⋅sin⁡2C=1−96⋅×34⇒a=3716∴(64a)2=1008

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