PrepRung
Sign inSign up free

JEE Advanced 2021 Paper 1 · Mathematics · Trigonometry and Inverse Trigonometric Functions · medium

In a triangle ABC the sides are BC=4, CA=5 and AB=6.

One or more options correct

In a triangle ABCABC the sides are BC=4BC=4, CA=5CA=5 and AB=6AB=6. Which of the following statements is (are) correct?
  1. A. ∠C=2∠A\angle C=2\angle A
  2. B. The inradius of the triangle is 73\dfrac{\sqrt7}{3}
  3. C. cos⁡B=58\cos B=\dfrac58
  4. D. The circumradius of the triangle is 47\dfrac{4}{\sqrt7}
Show answer and solution

Answer: A

Let a=4,b=5,c=6a=4,b=5,c=6. Then cos⁡A=b2+c2−a22bc=4560=34\cos A=\dfrac{b^2+c^2-a^2}{2bc}=\dfrac{45}{60}=\dfrac34 and cos⁡C=a2+b2−c22ab=540=18\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{5}{40}=\dfrac18. Since cos⁡2A=2⋅916−1=18=cos⁡C\cos2A=2\cdot\dfrac9{16}-1=\dfrac18=\cos C and both angles are acute enough to lie in (0,π)(0,\pi), C=2AC=2A: (A) true. The semi-perimeter is s=7.5s=7.5 and the area is 7.5⋅3.5⋅2.5⋅1.5=1574\sqrt{7.5\cdot3.5\cdot2.5\cdot1.5}=\dfrac{15\sqrt7}{4}, so r=Δs=72≠73r=\dfrac{\Delta}{s}=\dfrac{\sqrt7}{2}\ne\dfrac{\sqrt7}{3}: (B) false. cos⁡B=a2+c2−b22ac=2748=916≠58\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{27}{48}=\dfrac9{16}\ne\dfrac58: (C) false. R=abc4Δ=120157=87≠47R=\dfrac{abc}{4\Delta}=\dfrac{120}{15\sqrt7}=\dfrac{8}{\sqrt7}\ne\dfrac4{\sqrt7}: (D) false.
Report a mistake

Original question written in the JEE Advanced 2021 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