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

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

Question

(B) Given $5\cos A + 3 = 0$,so $\cos A = -\frac{3}{5}$.Since $A$ is an angle in a triangle and $\cos A < 0$,$A$ is an obtuse angle,so $\sin A = \sqrt{

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given $5\cos A + 3 = 0$,so $\cos A = -\frac{3}{5}$.Since $A$ is an angle in a triangle and $\cos A Then $	an A = \frac{\sin A}{\cos A} = \frac{4/5}{-3/5} = -\frac{4}{3}$.Let the roots be $\alpha = \sin A = \frac{4}{5}$ and $\beta = 	an A = -\frac{4}{3}$.Sum of roots: $\alpha + \beta = \frac{4}{5} - \frac{4}{3} = \frac{12 - 20}{15} = -\frac{8}{15}$.Product of roots: $\alpha \cdot \beta = (\frac{4}{5})(-\frac{4}{3}) = -\frac{16}{15}$.The quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - (-\frac{8}{15})x + (-\frac{16}{15}) = 0$.$x^2 + \frac{8}{15}x - \frac{16}{15} = 0$.Multiplying by $15$,we get $15x^2 + 8x - 16 = 0$.

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

Similar Questions

The number of all possible positive integral values of $\alpha$ for which the roots of the quadratic equation $6x^2 - 11x + \alpha = 0$ are rational numbers is:

If $x$ is real,then the interval in which no value of the expression $\frac{2(x^2+2x-11)}{2x-5}$ lies,is

The equation $\sqrt{3x^2 + x + 5} = x - 3$,where $x$ is real,has

What is the product of the real roots of the equation ${t^2}{x^2} + |x| + 9 = 0$?

If the equation $x^5-3x^4-5x^3+27x^2-32x+12=0$ has repeated roots,then the prime number that divides the non-repeated root of this equation is

Vedclass Test Series

Exam Paper Generator

Online Exam Module