trignometry

Asked by chiku999 | 30th Nov, 2009, 02:22: PM

Expert Answer:

sinA/sinB=p

sin2A/sin2B = p2

sin2A/(1 - cos2B) = p2

But cos2B = cos2A/q2

sin2A/(1 - cos2A/q2) = p2

q2sin2A/(q2 - cos2A) = p2

q2sin2A = p2(q2 - cos2A) = p2q2 - p+ p2sin2A

sin2A = (p2q2 - p2)/(q2-p2)

and hence cos2A = (q2 - p2q2)/(q2-p2)

tan2A = (p2q2 - p2)/(q2 - p2q2)

tan A = (±p/q)(q2-1)1/2/(1-p2)1/2

and we can see that tanA/tanB = p/q, hence

tan B = ±(q2-1)1/2/(1-p2)1/2

Regards,

Team,

TopperLearning.

Answered by  | 30th Nov, 2009, 03:26: PM

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