Question

If the vertices of a triangle $A,B,C$ are $A(0,0),B(2,1),C(9,-2)$ then $\cos B=$

• A. $\frac{16}{5\sqrt{17}}$
• B. $\frac{21}{\sqrt{290}}$
• C. $\frac{16}{5\sqrt{7}}$
• D. $\frac{-11}{\sqrt{290}}$

Answer 1

The correct option is D $\frac{-11}{\sqrt{290}}$

REF.Image.

using law of cosines; we get

$(AC{)}^2=(BC{)}^2+(AB{)}^2-2(BC\left)\right(AB)\cos B$

$85=58+5-2\sqrt{58}\sqrt{5}\cos B$

$85-63=-2\sqrt{58}\sqrt{5}\cos B$

$\frac{22}{2\sqrt{58}\sqrt{5}}=\cos B$

$\frac{-11}{\sqrt{58}\sqrt{5}}=\cos B$

$\therefore \cos B=\frac{-11}{\sqrt{290}}$