# Hard question involving trig and circles

• Mar 6th 2010, 05:55 PM
deltaxray
Hard question involving trig and circles
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 $\frac{\sqrt{6}}{4}$
• Mar 7th 2010, 05:43 AM
skeeter
Quote:

Originally Posted by deltaxray
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 $\frac{\sqrt{6}}{4}$

sketch in the radii to the chord endpoints.

let the endpoints be A and B, circle center O, and intersection of radius w/ chord be point C.

$AC = 2x$

$BC = x$

$OC = \frac{r}{2}
$

$\angle{ACO} = \theta$ is the acute angle between the chord and intersecting radius.

using the cosine law ...

$r^2 = (2x)^2 + \left(\frac{r}{2}\right)^2 - 2(2x)\left(\frac{r}{2}\right)\cos{\theta}$

$r^2 = 4x^2 + \frac{r^2}{4} - 2xr\cos{\theta}
$

$\textcolor{red}{\cos{\theta} = \frac{16x^2-3r^2}{8xr}}
$

now consider $\angle{OCB} = \phi$ , which is supplementary to $\theta$ .

$\cos{\phi} = -\cos{\theta}
$

using the cosine law again ...

$r^2 = x^2 + \left(\frac{r}{2}\right)^2 - 2(x)\left(\frac{r}{2}\right)\cos{\phi}$

$r^2 = x^2 + \left(\frac{r}{2}\right)^2 + 2(x)\left(\frac{r}{2}\right)\cos{\theta}$

$\textcolor{red}{\cos{\theta} = \frac{3r^2 -4x^2}{4xr}}
$

set the two expressions for $\cos{\theta}$ equal and solve for either x in terms of r or r in terms of x, then determine the value of $\cos{\theta}$ .