Friday, April 4, 2014

In a ΔABC, angle B = 90 degrees, AB = 6 units and AC = 10 units then the length of angle bisector AD from angle A is

 cosA=6/10=3/5 
A=arccos(3/5) - about 53.13 deg
so half is about A2=26.57 deg
therefore
cosA2=6/AD
cos26.57=6/AD
AD=6/cos26.57=6.71

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