Find the value of sin2A when(assuming 0<A<pi/2 ) . (a) cosA =15/17 (b)sinA=4/5 (c) tanA=5/12
Answers
Answered by
0
a)cos A=15/17
as we know sin^2+cos^2=1
sin^2+15^2/17^2=1
sin^2=1-15^2/17^2
take LCM17^2
=17^2-15^2/17^2
=289-225/17^2
=64/17^2
sinA×sinA =8×8/17×17
sinA=8/17 ;sin2A=8/17×2=16/17
b)sinA=4/5
sin2A=4/5×2=8/5
c)tanA=5/12
sinA/cosA=5/12
sinA=5
sin2A=2×5=10
as we know sin^2+cos^2=1
sin^2+15^2/17^2=1
sin^2=1-15^2/17^2
take LCM17^2
=17^2-15^2/17^2
=289-225/17^2
=64/17^2
sinA×sinA =8×8/17×17
sinA=8/17 ;sin2A=8/17×2=16/17
b)sinA=4/5
sin2A=4/5×2=8/5
c)tanA=5/12
sinA/cosA=5/12
sinA=5
sin2A=2×5=10
pulkitsharma035:
wrong
Similar questions