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Math Help - Hard question involving trig and circles

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    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}
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    Quote Originally Posted by deltaxray View Post
    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}<br />

    \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}<br />

    \textcolor{red}{\cos{\theta} = \frac{16x^2-3r^2}{8xr}}<br />

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

    \cos{\phi} = -\cos{\theta} <br />

    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}}<br />

    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} .
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