# Thread:

1. ## Circle

You have :
$<br /> OA=\sqrt{50},<br /> AB=6,<br /> BC=2<br />$
Calculate : OB
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2. Step 1. By Pythagoras' theorem, $AC = \sqrt{40}$. Let M be the midpoint of AC. Then $AM = \sqrt{10}$.

Step 2. OM is perpendicular to AC. By Pythagoras' theorem in the triangle OAM, $OM=\sqrt{40}$.

Step 3. Write $\alpha$ for the angle OAB, and $\beta$ for the angle BAC. From triangle OAM, $\cos(\alpha+\beta) = 1/\sqrt5$ and $\sin(\alpha+\beta) = 2/\sqrt5$. From triangle ABC, $\cos\beta = 3/\sqrt{10}$ and $\sin\beta = 1/\sqrt{10}$.

Step 4. Therefore $\cos\alpha = \cos\bigl((\alpha+\beta) - \beta\bigr) = 1/\sqrt2$ (using the trig formula for the cosine of the difference of two angles).

Step 5. Now apply the cosine rule in triangle OAB to find $OB = \sqrt{26}$.
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