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In a triangle abc, am is the median such that it divides angle a in 1:2. If am is...
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Let the angle bam=x.
Then angle mac=2x as given in the data.
Let angle bma = y
So angle amc = (180-y) since angle bma and angle bmc lies on the same line.
Since ad is a median bm=mc
abd is a triangle with ab perpendicular to bd.
Consider triangle bam. By law of sines we can write;
sinx/bm = siny/ab------(1)
Consider triangle amc. By law of sines we can write;
sin2x/mc = sin(180-y)/ac
We know that sin(180-y) = siny
sin2x/mc = siny/ac---------(2)
(sinx/sin2x)(mc/bm) = (siny/siny)(ac/ab)
It is given that bm=mc so mc/bm=1
Also sin2x = 2sinx*cosx
(sinx/(2sinx*cosx))(mc/bm) = (siny/siny)(ac/ab)
1/(2cosx) = ac/ab
ac = ab/(2cosx)
But ab = adcosx
ac = adcosx/(2cosx)
ac = ad/2
ad = 2ac
Posted by jeew-m on July 25, 2012 at 5:01 PM (Answer #1)
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