Question

If a angle $A$ of a $\Delta ABC$ satisfies the equation $5\cos A + 3 = 0$, then the roots of the quadratic equation, $9{x^2} + 27x + 20 = 0$ are,
(A) $\sin A,\sec A$
(B) $\sec A,\tan A$
(C) $\tan A,\cos A$
(D) $\sec A,\cot A$

Answer 1

Hint: Solve the given equation to find out the value of $\cos A$ and with the help of it, find all other trigonometric ratios, i.e., $\sin A,\tan A,CotA,\operatorname{Sec} A,\cos ecA$.

Complete step-by-step answer:
Given equation is: $5\cos A + 3 = 0$
We can solve it to find the value of $\cos A$.
$\therefore 5\cos A = - 3$
$ \Rightarrow \cos A = {{ - 3}}{5}$
Clearly, $\cos A$ lies in the second quadrant. Therefore , only $\sin A,\cos ecA$ are positive while all other trigonometric ratios are negative.
We know that, $\cos A = {{ - 3}}{5} = \dfrac{{Base}}{{Hypotenuse}}$
Now, we have to find the perpendicular of the triangle.
By applying Pythagoras theorem,
${\left( {Hypotenuse} \right)^2} = {\left( {Base} \right)^2} + {\left( {Perpendicular} \right)^2}$
$ \Rightarrow $${\left( 5 \right)^2} = {\left( 3 \right)^2} + {\left( {Perpendicular} \right)^2}$
$ \Rightarrow $${\left( {Perpendicular} \right)^2} = {\left( 5 \right)^2} - {\left( 3 \right)^2}$
$ \Rightarrow $${\left( {Perpendicular} \right)^2} = 25 - 9$
$ \Rightarrow $${\left( {Perpendicular} \right)^2} = 16$
$ \Rightarrow $$Perpendicular = \sqrt {16} $
$ \Rightarrow $$Perpendicular = 4$
Now, $\sin A = \dfrac{{Perpendicular}}{{Hypotenuse}} = {4}{5}$ and $\cos ecA = {1}{{\sin A}} = {5}{4}$
$\cos A = \dfrac{{Base}}{{Hypotenuse}} = \dfrac{{ - 3}}{5}$ and $\sec A = \dfrac{1}{{\cos A}} = \dfrac{{ - 5}}{3}$
$\tan A = \dfrac{{Perpendicular}}{{Base}} = \dfrac{{ - 4}}{3}$ and $\cot A = \dfrac{1}{{\tan A}} = \dfrac{{ - 3}}{4}$
Given quadratic equation is: $9{x^2} + 27x + 20 = 0$
To find the roots of given quadratic equation, compare it with $a{x^2} + bx + c = 0$ and we get-
$a = 9,b = 12,c = 27$
Now the roots of quadratic equation is calculated by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$
  x = \dfrac{{ - 27 \pm \sqrt {{{\left( {27} \right)}^2} - 4\left( 9 \right)\left( {20} \right)} }}{{2\left( 9 \right)}} \\
   \Rightarrow x = \dfrac{{ - 27 \pm \sqrt {729 - 720} }}{{18}} \\
   \Rightarrow x = \dfrac{{ - 27 \pm \sqrt 9 }}{{18}} \\
   \Rightarrow x = \dfrac{{ - 27 \pm 3}}{{18}} \\
    \\
 $
$ \Rightarrow $$x = \dfrac{{ - 27 + 3}}{{18}}$ and $x = \dfrac{{ - 27 - 3}}{{18}}$
$ \Rightarrow x = \dfrac{{ - 24}}{{18}}$ and $x = \dfrac{{ - 30}}{{18}}$
$ \Rightarrow x = \dfrac{{ - 4}}{3}$ and $x = \dfrac{{ - 5}}{3}$
We have, $\tan A = \dfrac{{ - 4}}{3}$ and $\sec A = \dfrac{{ - 5}}{3}$
So, the roots of the given equation are $\sec A,\tan A$.

Hence, option (B) is the correct answer.

Note: The another method to find out the roots of given quadratic equation is as follows:
$9{x^2} + 27x + 20 = 0$
$9{x^2} + \left( {15 + 12} \right)x + 20 = 0$
$9{x^2} + 15x + 12x + 20 = 0$
$3x\left( {3x + 5} \right) + 4\left( {3x + 5} \right) = 0$
$\left( {3x + 5} \right)\left( {3x + 4} \right) = 0$
$x = \dfrac{{ - 5}}{3}$ and $ \Rightarrow x = \dfrac{{ - 4}}{3}$