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enter image description here

In the figure above, $\angle MOP=\theta , \angle POP'=90^o$

$$\sin (90^o+\theta)=\sin \angle MOP'=\frac{M'P'}{OP'}(\text{how?})=\frac{OM}{OP}=\cos \theta$$

Sine is opposite side/hypotenuse, then in this case how does the author determine the sine of an obtuse angle? (Is it possible with elementary geometry? This topic is far before laws of sine and cosine)

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  • $\begingroup$ my query is how $\sin MOP'=\frac{M'P'}{OP'}$ ? $\endgroup$ – Vikram Jun 16 '15 at 15:00
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This depends on your definition of $\sin$. It is not abnormal to define $\sin$ so that your equation holds. So either it is by definition, or you probably have to use some simple formula, like:

the formula $\sin (180 - x) = \sin(x)$. So $$ \frac{M'P'}{OP'} = \sin M'OP' = \sin (180 - M'OP') = \sin(90 + \theta). $$

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  • $\begingroup$ "sin(180-x)=sin x" is after this topic, so I can not use it $\endgroup$ – Vikram Jun 16 '15 at 15:03
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    $\begingroup$ @Vikram But in this case you can just refer to the picture and see that it's true. $\endgroup$ – gebruiker Jun 16 '15 at 15:05
  • $\begingroup$ Oh, I feel you both are right, thanx $\endgroup$ – Vikram Jun 16 '15 at 15:08

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