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Let $OAB$ be a triangle with $\angle AOB = \theta, \angle OBA = \alpha$ and $|OA| = r$. Then show $$\tan(\alpha)= \frac{r\sin(\theta)}{a-r\cos(\theta)}.$$

enter image description here

I'm not sure how this is done. There's a similar question on Stack Exchange, but I don't understand converting the image to a parallelogram to find tan. I thought to maybe use the Sine Rule, but I'm not sure if that works out? I'm not the best at trigonometry.

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Drop perpendicular from $A$ on $OB$. Name this point $P$. We will now get $OP = r \cos \theta$ and $AP = r \sin \theta$. Also $OB = a$

$$\tan\alpha = \frac{AP}{BP} = \frac{AP}{OB-OP} = \frac{r \sin \theta}{a-r \cos \theta}$$

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Drop a perpendicular from $A$ to $OB$, let that intersect $OB$ at point $P$

In right $\triangle APB$, $\tan \alpha = \frac{AP}{PB}$

In right $\triangle OPA$, $AP = r\sin\theta$ and $OP = r\cos\theta$

$PB = OB - OP = a - r\cos\theta$

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  • 1
    $\begingroup$ Why the downvote? $\endgroup$ – Deepak Oct 29 '17 at 5:08

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