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QJEEM18Ma98d2: JEE Main 2018 Mathematics - Class 11 JEEM18Ma98d2 Solution: If an angle A of a ABC satiesfies 5 cosA + 3 = 0, then the…

If an angle A of a Δ\Delta ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x2 + 27x + 20 = 0 are :

Multiple Choice Options

  • A secA, cotA
  • B sinA, secA
  • C secA, tanA
  • D tanA, cosA

Step-by-Step Verified Solution

Here, 9x2 + 27x + 20 = 0

∴   \therefore\,\,\, x = −b±b2−4ac2a{{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}

⇒\Rightarrow    \,\,\, x = −27±272−4×9×202×9{{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}

⇒\Rightarrow    \,\,\, x = −- 43{4 \over 3} , −- 53{5 \over 3}

Given, cosA = −- 35{3 \over 5}

∴   \therefore\,\,\, sec A = 1cos⁡A{1 \over {\cos A}} = −- 53{5 \over 3}

Here, A is an obtuse angle.

∴   \therefore\,\,\, tan A = −- sec⁡2A−1=−43.\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.

Hence, roots of the equation are sec A and tan A.

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