 My Math Forum Line length from mid-point circle to outside diameter circle

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 March 29th, 2017, 11:12 AM #1 Newbie   Joined: Mar 2017 From: Netherlands Posts: 1 Thanks: 0 Line length from mid-point circle to outside diameter circle I might be stupid, but I have been looking at this for a few hours: either I am stupid or it is unsolvable (which could be an option). I have two circles, one with a diameter of 16mm, the other one with diameter of 32mm. A line which starts at the mid-point of the 32mm circle, connects to the outside of the 16mm circle. The angle it makes is 20 degrees. How should I calculate the length of this line? I tried a lot of triangle equations but I did not manage to get the answer... I do not need the answer, I need the calculation how its done actually. If someone knows how to do it, please tell me how I should do it, you do not have to write everything down if you do not want to. Thanks! March 29th, 2017, 03:06 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2336 The greatest value angle a can have is approximately 19.47122°, so your configuration is impossible. Thanks from topsquark March 29th, 2017, 03:26 PM #3 Senior Member   Joined: Jul 2008 From: Western Canada Posts: 5,199 Thanks: 47 Well, first of all there's a problem with your diagram. A line from the centre of the larger circle that's tangent to the smaller one makes an angle of only 19.47° with the line between centres. So, your angle of 20° is impossible. The line would never touch the smaller circle. Oops. Skipjack beat me to it. Thanks from topsquark March 29th, 2017, 04:56 PM   #4
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Joined: Jul 2008

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Thanks: 47

Here is a general solution (refer to attached diagram) when the angle a is small enough for a valid solution. The things we know are shown in green.

We can first solve for the unknown angles b and c using the law of sines:

$\dfrac{\sin(a)}{A}=\dfrac{\sin(b)}{B}=\dfrac{\sin (c)}{C}$

Since we know the lengths A and B and angle a, we can use this formula to solve for angle b:

$\dfrac{\sin(a)}{A}=\dfrac{\sin(b)}{B}$

Rearranging:

$b=\arcsin(\frac{B}{A}\sin(a))$

We have to be careful here though. The arcsin function has two possible results, an acute angle and an obtuse angle. This function will normally return the smaller value. We can see from the diagram, that angle b is obtuse, and so we have to subtract the arcsin value from 180°, thus:

$b=180-\arcsin(\frac{B}{A}\sin(a))$

Now that we know angles a and b, we can use the fact that the included angles of a triangle add up to 180°. Therefore:
$c=180-b-a$

We now know all of the angles and two sides A and B. We can now apply the law of cosines to find the unknown side C. The law of cosines is:

$C^2=A^2+B^2-2AB\cos(c)$

Therefore:

$C=\sqrt{A^2+B^2-2AB\cos(c)}$
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Attached Images Geometry01.jpg (5.9 KB, 14 views)

Last edited by Yooklid; March 29th, 2017 at 05:00 PM. Tags circle, diameter, length, line, midpoint Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mark eaton24 Applied Math 15 January 19th, 2016 04:09 AM Ash Algebra 1 March 12th, 2014 10:50 AM mathkid Calculus 6 October 1st, 2012 07:49 PM soulrain Algebra 7 January 5th, 2012 08:51 PM mad Algebra 3 August 18th, 2010 01:20 PM

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