A radius of a circle divides a chord in the ratio 2:1 and is bisected by the chord. Show that the cosine of the angle between the radius and the chord is (1/4)*√6.
Thanks guys for any help
A radius of a circle divides a chord in the ratio 2:1 and is bisected by the chord. Show that the cosine of the angle between the radius and the chord is (1/4)*√6.
Thanks guys for any help
make a sketch. let the chord be divided into lengths 2x and x. connect radii to the endpoints of the chord. two triangles are formed.
using the law of cosines ...
first triangle with the acute angle at the intersection of the given radius and chord:
$\displaystyle r^2 = (2x)^2 + \left(\frac{r}{2}\right)^2 - 2(2x)\left(\frac{r}{2}\right)\cos{\theta}$
simplifying ...
$\displaystyle \frac{3r^2}{4} = 4x^2 - 2rx\cos{\theta}$
second triangle with the obtuse angle at the same intersection:
$\displaystyle r^2 = x^2 + \left(\frac{r}{2}\right)^2 - 2x\left(\frac{r}{2}\right)\cos{(180 - \theta)}$
simplifying ...
$\displaystyle \frac{3r^2}{4} = x^2 + rx\cos{\theta}$
use the two simplified equations to solve for
$\displaystyle \cos{\theta}$
I'd start by solving for $\displaystyle \cos{\theta}$ in terms of r and x.