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Say there are two lines that are tangent to a circle at points A and B, so that they intersect at an external point C, as shown below.

Two congruent right triangles are formed, since the tangent line is perpendicular to the radius. So line AC has the same length as line BC.

If the angle ACB is known, how can the length of line AC be calculated?

For example, if the radius $r$ is 1, and the angle ACB is $60\unicode{xb0}$, what is the length of line AC?

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    $\begingroup$ Note that ACO (O is the circle centre) is half an equilateral triangle and that $\angle ACO=\frac12\angle ACB$. Apply trigonometric formulas and you're done. $\endgroup$ – Parcly Taxel Sep 29 '16 at 0:28
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    $\begingroup$ You can use the tangent of the angle inside the right triangles and $r$ (one of the legs) to find $AC$. If you know $m\angle ACB$ then you know the two angles of the right triangle. $\endgroup$ – Jared Sep 29 '16 at 0:31
  • $\begingroup$ That's a good point, it's simpler than I thought. So if we let $l = AC$ and $\theta = ACB$ , then the relation is $l = r ÷ \tan{(\theta/2)}$ $\endgroup$ – Vermillion Sep 29 '16 at 0:38
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If $\angle ACB=60$ then $\angle ACO=30$ and therefore $OC=2$. Now you can use pythagoras and find $AC$. Using pythagoras : $OC^2=OA^2+AC^2$ and you will have: $4=1+AC^2$ Therefore $AC=\sqrt 3$

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