In triangle ABC,the value of angle A is obtai | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $3 \cos A + 2 = 0$,we have $\cos A = -\frac{2}{3}$.Since $A$ is an angle in a triangle,$\sin A > 0$. Thus,$\sin A = \sqrt{1 - \cos^2 A} = \s

Vedclass · Accepted Answer

$6 x^2 + \sqrt{5} x - 5 = 0$

Vedclass · Answer

(C) Given $3 \cos A + 2 = 0$,we have $\cos A = -\frac{2}{3}$.Since $A$ is an angle in a triangle,$\sin A > 0$. Thus,$\sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.Then,$	an A = \frac{\sin A}{\cos A} = \frac{\sqrt{5}/3}{-2/3} = -\frac{\sqrt{5}}{2}$.The quadratic equation with roots $\alpha$ and $\beta$ is $x^2 - (\alpha + \beta)x + \alpha\beta = 0$.Here,$\alpha = \sin A = \frac{\sqrt{5}}{3}$ and $\beta = 	an A = -\frac{\sqrt{5}}{2}$.Sum of roots: $\alpha + \beta = \frac{\sqrt{5}}{3} - \frac{\sqrt{5}}{2} = \frac{2\sqrt{5} - 3\sqrt{5}}{6} = -\frac{\sqrt{5}}{6}$.Product of roots: $\alpha\beta = \left(\frac{\sqrt{5}}{3}ight) \left(-\frac{\sqrt{5}}{2}ight) = -\frac{5}{6}$.The equation is $x^2 - (-\frac{\sqrt{5}}{6})x - \frac{5}{6} = 0$,which simplifies to $x^2 + \frac{\sqrt{5}}{6}x - \frac{5}{6} = 0$.Multiplying by $6$,we get $6x^2 + \sqrt{5}x - 5 = 0$.

In $\triangle ABC$,the value of $\angle A$ is obtained from the equation $3 \cos A + 2 = 0$. The quadratic equation,whose roots are $\sin A$ and $\tan A$,is

Similar Questions

$\alpha, \beta, \gamma, 2, \varepsilon$ are the roots of the equation $x^5+4 x^4-13 x^3-52 x^2+36 x+144=0$. If $\alpha < \beta < \gamma < 2 < \varepsilon$,then find the value of $\alpha+2 \beta+3 \gamma+5 \varepsilon$.

The values of $a$ for which the expression $(a^2 - 1)x^2 + 2(a - 1)x + 2$ is positive for all $x \in \mathbb{R}$ are:

When the roots of $x^3+\alpha x^2+\beta x+6=0$ are increased by $1$,if one of the resultant values is the least root of $x^4-6 x^3+11 x^2-6 x=0$,then

The roots of the equation $|x - 2|^2 + |x - 2| - 6 = 0$ are:

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+4x^2-9x-36=0$ and $\alpha < \beta < \gamma$,then find the value of $\alpha+2\beta+3\gamma$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module