Question

If $a,b,c$ are the sides of $△ABC$ and $\sin \theta$ and $\cos \theta$ are the roots of equation $a{x}^2-bx+c=0$,

then $\cos B$ equals to

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

Answer 1

The correct option is B $\frac{c}{2a}-1$

We have, $\sin \theta +\cos \theta =\frac{b}{a};\sin \theta \cos \theta =\frac{c}{a}$

$\therefore {(\sin \theta +\cos \theta )}^2=1+2\sin \theta \cos \theta =1+2\frac{c}{a}$

$\Rightarrow \frac{b^2}{a^2}=1+2\frac{c}{a}\Rightarrow b^2=a^2+2ac\Rightarrow b^2-a^2=2ac$

Now,

$\cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{c^2-(b^2-a^2)}{2ac}$

$=\frac{c^2-2ac}{2ac}=(\frac{c}{2a}-1)$