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Math Help - Circle

  1. #1
    Super Member dhiab's Avatar
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    Circle

    You have :
     <br />
OA=\sqrt{50},<br />
AB=6,<br />
BC=2<br />
    Calculate : OB
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    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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