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Three points at coordinates $(0,c)$, $(p,q)$, $(0,d)$ respectively. The angle at $(p,q)$ between $(0,c)$ and $(0,d)$ is $θ$. Find $d$.

P.s. This isn't homework.

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1 Answer 1

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Hint :

Let $A(0,c)$, $B(0,d)$, and $C(p,q)$. Length $AB=c-d$, \begin{align} AC^2=p^2+(c-q)^2 \end{align} and \begin{align} BC^2=p^2+(q-d)^2 \end{align} Using cosine formula, we get \begin{align} AB^2&=AC^2+BC^2-2\cdot AC\cdot AB\cos\theta\\ (c-d)^2&=p^2+(c-q)^2+p^2+(q-d)^2-2\sqrt{(p^2+(q-d)^2)(p^2+(q-d)^2)}\cos\theta\\ \end{align}

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  • $\begingroup$ Should be \begin{align} AB^2&=AC^2+BC^2-2\cdot AC\cdot BC\cos\theta\\ (c-d)^2&=p^2+(c-q)^2+p^2+(q-d)^2-2\sqrt{(p^2+(c-q)^2)(p^2+(q-d)^2)}\cos\theta\\ \end{align} $\endgroup$
    – Zimul8r
    May 1, 2014 at 18:10

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