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0|0|0|||||Ptolemy%27s Theorem and Cosines - How Does This Work%3F|matthewjheaney||16:56:01|11/15/2009|On the page here%3A%0D%0A%0D%0Ahttp%3A%2F%2Fwww.cut-the-knot.org%2Fproofs%2Fsine_cosine.shtml%0D%0A%0D%0Aa quadrilateral is inscribed in a circle having diameter 1.%0D%0A%0D%0AThe page lists some equivalences %28down towards the bottom%29 between the angles alpha and beta and the length of the sides of the quadrilateral.%0D%0A%0D%0AI understand from the Sine Rule that sin%28alpha%29 %3D AC%2C and sin%28beta%29 %3D DC.%0D%0A%0D%0AWhat I do not understand is how the equivalences cos%28alpha%29 %3D AB and cos%28beta%29 %3D BD were derived.  Did a step go missing there somewhere%3F  I%27m scratching my head trying to prove it.%0D%0A%0D%0AThanks%2C%0D%0AMatt
2|1|0|||||RE%3A Ptolemy%27s Theorem and Cosines - How Does This Work%3F|alexb||17:56:48|11/15/2009|%3EI understand from the Sine Rule that sin%28alpha%29 %3D AC%2C and %0D%0A%3Esin%28beta%29 %3D DC. %0D%0A%0D%0APerhaps the apppeal to the Sine Law is misleading. Sorry for that. The simple truth is that both sine and cosine are originally defined in a right triangle as%2C respectively%2C the ratio of the opposite leg to the hypotenuse and the ratio of the adjacent leg to the hypotenuse. In triangle ABC with the hypotenuse BC %28since it%27s a diameter%2C angle A is right%29%2C sin%28%26alpha%3B%29 %3D AC%2FBC and cos%28%26alpha%3B%29 %3D AB%2FBC. Now%2C because%2C by our choice BC %3D 1%2C sin%28%26alpha%3B%29 %3D AC and cos%28%26alpha%3B%29 %3D AB.%0D%0A