tag:blogger.com,1999:blog-7320798620465608376.post6193720891072303039..comments2016-05-05T17:44:30.415-07:00Comments on A bit of maths: Maths puzzle: a bit of trigPhilip Kinlenhttp://www.blogger.com/profile/14448358012402515803noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-7320798620465608376.post-9534676702612020222014-05-07T02:33:18.244-07:002014-05-07T02:33:18.244-07:00The answer is sqrt(6):
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Pr...The answer is sqrt(6):<br />________________________<br />Proof 1:<br /><br />Use cartesian co-ordinates:<br /><br />let:<br />P = (0,0)<br />B= ( 3, 0) , PAB is isosceles<br />A = ( 3 cos x, 3 sin x ) , where x is the angle at APB<br />C = ( 3 + 2 cos x, 2 sin x )<br />Q = ( 3 , r ) , where Q is the center of the circle and r is the circle radius, <br /><br />r ^ 2 = | QC | ^ 2<br /> = 4 + r^2 - 4 r sin x<br />so r = 1/ sin x (i)<br /><br />and looking at BPQ we see that:<br /> tan( x / 2 ) = r / 3 (ii)<br /><br />so, combining (i) and (ii) : <br /> sin x / ( 1 + cos x ) = 1 / ( 3 * sin x )<br /><br />thus cos x = 2 /3<br /><br />and so | AC | = sqrt (6)<br /><br />________________________<br />Proof 2:<br />Outline:<br />By looking at the angles round the centre of the circle we can show that the triangles PAB and ABC are similar isoscelese triangles.<br />( that takes a bit of work).<br /><br />Once we know that we look at the ratio of the lengths, we now know that:<br /> 3/L = L/2 <br />where L = |AC| = |AB| <br /><br />hence |AC| = sqrt(6) <br /><br /><br />Philip Kinlenhttps://www.blogger.com/profile/14448358012402515803noreply@blogger.com