Thread: Hard question involving trig and circles

1. 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 $\displaystyle \frac{\sqrt{6}}{4}$

2. 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 $\displaystyle \frac{\sqrt{6}}{4}$
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.

$\displaystyle AC = 2x$

$\displaystyle BC = x$

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

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

using the cosine law ...

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

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

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

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

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

using the cosine law again ...

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

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

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

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