In a triangle ABC the value of A is given by 5 A + 3 = 0, then th… - Vidyadip

MCQ

In a triangle $ABC$ the value of $\angle A$ is given by $5\cos A + 3 = 0$, then the equation whose roots are $\sin A$ and $\tan A$ will be

A
$15{x^2} - 8x + 16 = 0$
✓
$15{x^2} + 8x - 16 = 0$
C
$15{x^2} - 8\sqrt 2 x + 16 = 0$
D
$15{x^2} - 8x - 16 = 0$

✓

Answer

Correct option: B.

$15{x^2} + 8x - 16 = 0$

b
(b) Given that $5\cos A + 3 = 0$ or $\cos A = - \frac{3}{5}$

Let $\alpha = \sin A$and $\beta = \tan A$, then the sum of roots $ = \alpha + \beta = \sin A + \tan A$

$ = \sin A + \frac{{\sin A}}{{\cos A}} = \frac{{\sin A}}{{\cos A}}(1 + \cos A)$

$ = \frac{{\sqrt {1 - 9/25} }}{{ - 3/5}}\left( {1 - \frac{3}{5}} \right) = \frac{4}{{ - 5}}.\frac{5}{3}.\frac{2}{5} = \frac{8}{{ - 15}}$

and product of roots $\alpha .\beta = \sin A\tan A = \frac{{{{\sin }^2}A}}{{\cos A}}$

$ = \frac{{16/25}}{{ - 3/5}} = - \frac{{16}}{{25}} \times \frac{5}{3} = - \frac{{16}}{{15}}$

Thus required equation is ${x^2} - (\alpha + \beta )x + \alpha \beta = 0$

==> ${x^2} + \frac{{8x}}{{15}} - \frac{{16}}{{15}} = 0$

==> $15{x^2} + 8x - 16 = 0$

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