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A circle has diameter $AD$ of length $400$.

$B$ and $C$ are points on the same arc of AD such that $|AB|=|BC|=60$.

What is the length $|CD|$?

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Here we go again, taken from – Erick Wong Apr 16 '13 at 18:02

We describe a trigonometric approach. We will use a couple of times the following fact. Let $\triangle PQR$ be isosceles, with $PQ=PR=a$ and $QR=b$. Let $y=\angle QPR$. Then $\frac{b/2}{a}=\sin(y/2)$.

This is easily proved by joining $P$ to the midpoint $M$ of $QR$.

Now let $O$ be the centre of the circle, and let $\angle AOB=x$. Then $\angle BOC=x$. and $\angle COD=180^\circ-2x$. It follows that half of $\angle COD$ is $90^\circ -x$.

By basic trigonometry, the length of $CD$ is $200\sin(90^\circ-x)$, that is, $200\cos x$.

It rmains to find $\cos x$. From $\triangle AOB$, we see that $\sin(x/2)=\frac{30}{200}=\frac{3}{20}$.

But by a double angle identity, we have $$\cos x=1-2\sin^2(x/2),$$ and now we know $\cos x$.

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We can get around evaluating the angle $x$ itself or stating the value of $\cos x$, since the law of cosines gives us $60^2 = 2 \cdot 200^2 \cdot (1 - \cos x)$, so we can express the length of CD entirely in terms of either the length of AB or BC and the radius of the circle. (Otherwise the method I was going to propose is related to the one Mr. Nicolas gives.) – RecklessReckoner Apr 15 '13 at 5:41
Nice alternative. I did an "automatic" crude calculation. Actually, the real calculation I wanted to do used analytic geometry, but I got lazy, the idea is simple but the typing less so. – André Nicolas Apr 15 '13 at 5:45
There are a few approaches possible, but the trigonometric one is probably the most direct; analytic geometry would indeed take more writing. The route I took involves more trig identities, but leads to the result that the length of CD is $2 \cdot R \cdot \cos \theta$, where $R$ is the radius of the circle and $\theta$ is the angle subtended by either AB or BC. – RecklessReckoner Apr 15 '13 at 5:50

Here's one approach...

Given AD = 400, AB = BC = 60

Now, ABD and ACD are right triangles in semicircle ABCD. Using the Pythagoras theorem, BD² = AD² − AB², BD² = 400² − 60² = 156400

Similarly, CD² = AD² − AC² CD² = 400² − AC² ———— (1)

ABCD is a concyclic. Thus, from Ptolemy’s theorem, (AC)(BD) = (AB)(CD) + (AD)(BC) 20√391 AC = 60CD + 400(60), 20√391 AC = 60CD + 24000 —————- (2)

Solving for positive value of CD from (1) and (2) , and getting : CD = 382

Anyway I shouldn't be giving away direct answers to you....

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the simplest trig approach? (minimal concepts)

let point $P$ bisect chord AB, then $\angle AOP$ is $ sin^{-1}(\frac{30}{200})$, and as $\triangle AOB$ and $\triangle BOC$ are equivalent the $\angle AOC$ is just 4 times this. thus $\angle COD$ is $\pi - (4*\angle AOP)$ radians.

let point $Q$ bisect chord CD, then $\angle COQ$ is half $\angle COD$. for $\triangle COQ$, length |CQ|, opposite our $\angle COQ$, is $200 * sin(\angle COQ)$. we have $|CQ|$ and $|CQ| = \frac{1}{2}|CD|$

so in a (oversized) nutshell...

$$|CD| = 2 * 200 * sin(\frac{\pi~-~4 * sin^{-1}(\frac{30}{200})}{2}) $$

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