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JEE Advanced 2019 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

In an acute-angled triangle ABC, tan A=2 and tan B=3, and the circumradius is R=√(10). Let a,b,c be the sides opposite to A,B,C.

One or more options correct

In an acute-angled triangle ABCABC, tan⁡A=2\tan A=2 and tan⁡B=3\tan B=3, and the circumradius is R=10R=\sqrt{10}. Let a,b,ca,b,c be the sides opposite to A,B,CA,B,C. Which of the following statements is/are correct?
  1. A. C=45∘C=45^\circ
  2. B. The area of the triangle is 1212
  3. C. The shortest side has length 44
  4. D. sin⁡2A+sin⁡2B+sin⁡2C=125\sin^2A+\sin^2B+\sin^2C=\dfrac{12}{5}
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Answer: A, B

Since A+B+C=πA+B+C=\pi, tan⁡C=−tan⁡(A+B)=−2+31−6=1\tan C=-\tan(A+B)=-\frac{2+3}{1-6}=1, so C=45∘C=45^\circ and (A) is correct. With tan⁡A=2\tan A=2, sin⁡A=25\sin A=\frac2{\sqrt5}; with tan⁡B=3\tan B=3, sin⁡B=310\sin B=\frac3{\sqrt{10}}; and sin⁡C=12\sin C=\frac1{\sqrt2}. Sides are 2Rsin⁡2R\sin of the angles: a=42a=4\sqrt2, b=6b=6, c=25c=2\sqrt5. The area is 2R2sin⁡Asin⁡Bsin⁡C=2⋅10⋅35=122R^2\sin A\sin B\sin C=2\cdot10\cdot\frac35=12, so (B) is correct. The shortest side is c=25≈4.47c=2\sqrt5\approx4.47, not 44, so (C) is wrong. Finally sin⁡2A+sin⁡2B+sin⁡2C=45+910+12=115\sin^2A+\sin^2B+\sin^2C=\frac45+\frac9{10}+\frac12=\frac{11}5, not 125\frac{12}5, so (D) is wrong.
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Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake