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2014-10-16T23:34:05+05:30

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SinAcosB - sinBCosA = 1/2...........i

cosAcosB - sinAsinB = 1/2   .........ii

so  i divide ii
 sinAcosB - sinBCosA  = cosAcosB - sinAsinB
sinA(cosB + sinB) = CosA(sinB + CosB)
so tanA = 1
   A = 45°

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thanx a lot :)
is the answer correct
plz mark as best
only if i have 2 ans. i can mark urs as best
n btw thanx again
2014-10-17T02:14:19+05:30
Let sinAcosB - sinBCosA = 1/2 be equation (i) and
cosAcosB - sinAsinB = 1/2 be equation (ii)
Hence by dividing equation two by one we get;
sinAcosB - sinBCosA  = cosAcosB - sinAsinB
sinA(cosB + sinB) = CosA(sinB + CosB)
so tanA = 1
   A = 45°
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