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On the image below line DB is tangent to the circle. The problem is to try and express both $\Delta OBC$ and $\Delta OBD$ in terms of $\sin\theta$ and $\cos\theta$.

For $\Delta OBC$ $$A=\frac{1}{2}OB(CA)$$ I found out all the trig functions for it$$\sin\theta=\frac{CA}{OC}$$ $$\cos\theta=\frac{OA}{OC}$$ $$\tan\theta=\frac{CA}{OA}$$ I am having trouble using the trig functions to express area in $\sin\theta$ and $\cos\theta$

enter image description here

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    $\begingroup$ Area of triangle $OBD$ is $\frac{1}{2} \cdot OB \cdot DB = \frac{1}{2} \tan \theta $ $\endgroup$ – TheSimpliFire Oct 30 '17 at 20:29
  • $\begingroup$ @TheSimpliFire That clears things up, thanks $\endgroup$ – RedRobin Oct 30 '17 at 21:36
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    $\begingroup$ $\tan\theta=DB$, then use similar triangles. $\endgroup$ – amd Oct 30 '17 at 22:36
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$\Delta OBC$ and $\Delta OBD$ are not similar triangles. $\Delta OBC$ is not a right triangle.

Also, by "$\theta$", I assume you mean $x$? If so, $\tan\theta\neq\frac{CA}{OC}$, it is $\tan\theta=\frac{CA}{OA}$. Because $OC = 1$, the trig functions simplify to $$\sin \theta = CA $$ $$\cos \theta = OA$$

$$A_{OBC}= \frac{1}{2}(OB)(CA) = \frac{1}{2}(1)(\sin \theta) = \frac{1}{2}\sin \theta $$

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