I need to find the angle at origin caused by two lines (one is the radius of one circle, and the other is tangent to the other circle).

Please see image below:

The Point A on the green circle is known, the dashed line and shorter full line begin at the origin. The other line is tangent to the red circle and passing through Point A.

How would I calculate the angle a?

I know this is probably a basic question but my head is burnt out at work and I just need reassurance before making an epic mistake.

  • 1
    $\begingroup$ The radius of the big circle is $\sqrt{2}$ but the radius $r$ of a small circle is unknown. The angle $\alpha$ depends on $r.$ $\endgroup$ – kmitov Mar 13 '14 at 10:24
  • $\begingroup$ The radius of the red circle...and also: are both circles centered at the origin? $\endgroup$ – DonAntonio Mar 13 '14 at 10:50

Since from the drawing it is clear that $\;r<\sqrt 2\;$, we have that if $\;\alpha\;$ is the wanted angle then

$$\cos\alpha=\frac r{\sqrt2}\implies \alpha=\arccos\frac r{\sqrt2}$$

  • $\begingroup$ to find the angle a do we simply do a = arccos(radius of small circle/radius of larger circle) $\endgroup$ – Amie-lea Mar 13 '14 at 11:10
  • $\begingroup$ @Amie-lea Well, yes...choosing accordingly the range of the angle. Isn't the above basic trigonometric formula clear enough? $\endgroup$ – DonAntonio Mar 13 '14 at 11:12
  • $\begingroup$ Yes I just wanted to clarify it as the point (1,1) was just added for simplicity. $\endgroup$ – Amie-lea Mar 13 '14 at 11:16
  • $\begingroup$ In fact the point $\;(1,1)\;$ has no importance whatsoever in this case, @Amie-lea : going by the same drawing, any other point on the outer circle gives the very same result. $\endgroup$ – DonAntonio Mar 13 '14 at 11:17
  • $\begingroup$ The same result with sqrt(2) on the bottom, or would this be replaced with the radius of the outer circle in other circumstances? Sorry for basic questions just want clarification $\endgroup$ – Amie-lea Mar 13 '14 at 11:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.