Question

If $\sin \theta$ and $-\cos \theta$ are the roots of the equation $a{x}^2-bx-c=0$, where a, b, and c are the sides of a triangle ABC, then cos B is equal to

• A. $1-\frac{c}{2a}$
• B. $1-\frac{c}{a}$
• C. $1+\frac{c}{2a}$
• D. $1+\frac{c}{3a}$

Answer 1

The correct option is C $1+\frac{c}{2a}$

If $\sin \theta$ and $-\cos \theta$ are the roots of the eqution $a{x}^2-bx-c=0$
Then $\sin \theta -\cos \theta =b/a$........(1)
And $\sin \theta (-\cos \theta )=-c/a\Rightarrow \sin \theta \cos \theta =c/a$ .....(2)
Using $(1{)}^2+2\cdot (2)$
$\Rightarrow {\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta +2\sin \theta \cos \theta =b^2/a^2+2c/a$
$\Rightarrow 1=b^2/a^2+2c/a\Rightarrow a^2=b^2+2ac$ ....(3)
Thus $\cos B=\frac{c^2+a^2-b^2}{2ac}=\frac{c^2+b^2+2ac-b^2}{2ac}$ using (3)
$=\frac{c^2+2ac}{2ac}=1+\frac{c}{2a}$