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Math Help - trigo problem no. 12

  1. #1
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    trigo problem no. 12

    p202 q20 re.ex.9
    don't know how to do [b][ii]---show that ABC is equilaqteral. thanks
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  2. #2
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    Hello, afeasfaerw23231233!

    I've made one observation . . . \Delta ODE is equilateral.
    . . But I haven't gone any further.

    We have these areas: . \begin{Bmatrix}\Delta CED & = & \frac{1}{2}ab\cos^2\theta\sin\theta \\ \text{quad }ADEB & = & \frac{1}{2}ab\sin^3\!\theta \end{Bmatrix}

    We are given the ratio of these areas: . \frac{\Delta CED }{\text{quad }ADEB } \;=\;\frac{\frac{1}{2}ab\cos^2\!\theta\sin\theta}{  \frac{1}{2}ab\sin^3\!\theta} \;=\;\frac{1}{3}

    . . which simplifies to: . \frac{\cos^2\!\theta}{\sin^2\!\theta} \:=\:\frac{1}{3}\quad\Rightarrow\quad\tan^2\!\thet  a \:=\:3\quad\Rightarrow\quad \tan\theta \:=\:\sqrt{3}

    Hence: . \theta \:=\:60^o

    Since \Delta ODE is isosceles (OD = OE = \text{radius}),
    . . then: . \angle DEO = 60^o . . . Hence: . \angle DOE = 60^o

    Therefore: . \Delta ODE is equilateral.

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  3. #3
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    sorry, i made a calculating mistake, theta is not = 45 degree!
    but i still cannot prove triangle ABC is equilateral
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    Last edited by afeasfaerw23231233; February 1st 2008 at 09:40 PM.
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