# Intersection point of a tangent of a circle with a line

So, I'm studying math (for school) and I'm having some trouble in this one below:

01.13 - In the Picture below, the s line crosses the P dot and by the center of the circle of R radius, intersecting by the Q dot, between P and the center. Besides, the t line crosses by P, it's tangent by the circle and creates an α angle with the line s. If PQ = 2R, so the α cosine iquals:

$$\frac{ {2}{\sqrt{2}}} {3}$$

How can I get there?

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Join the centre (C) and the tangent point (Y).

The distance $PC = PQ + QC = 2r + r = 3r$

The distance from Y to C is r

Thus,

\begin{align} \sin(\alpha) &= \dfrac{r}{3r} \\ &=\dfrac{1}{3}\\ \cos(\alpha) &= \sqrt{(1-\sin^2(\alpha)}\\ &=\sqrt{1-\dfrac{1}{9}}\\ &=\dfrac{2\sqrt 2}{3} \end{align}

EDIT : Since you were not aware of $$\sin^2\theta + \cos^2 \theta = 1$$

\begin{align} \cos(\alpha)&=\dfrac{PY}{PC}\\ \text {By Pythagoras Theorem,}\\ PC^2&=PY^2+CY^2\\ (3r)^2&=PY^2+r^2\\ PY^2&=8r^2\\ \implies \cos(\alpha)&=\dfrac{2\sqrt 2 r}{3r}\\ \cos(\alpha)&=\dfrac{2\sqrt 2}{3}\\ \end{align} We apply pythagoras because the tangent is perpendicular to the radius at the point of contact. That's a theorem.

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Thanks for answering, I'm reading your answer a couple of times to understand it, as english isn't my first language I didn't understand it by the first time. thank you so much for the quick answer! –  Henrique Foletto Feb 23 '13 at 22:04
@HenriqueFoletto. My pleasure. Let me know if you need clarification on any step. –  Inquest Feb 23 '13 at 22:05
I just didn't understand yet how you transformed the sin into a cos, is it a regulation (I don't know if it's the right word)? Thanks again. –  Henrique Foletto Feb 23 '13 at 22:07
Do you know the pythagorean identity $sin(x)^2 + cos(x)^2 = 1$ ? –  user38034 Feb 23 '13 at 22:12
No, I didn't know that. Thank you so much for that info. –  Henrique Foletto Feb 23 '13 at 22:15