sin2A =12/13 and A is in the first quadrant sin 2A?

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sin2A =12/13 and A is in the first quadrant sin 2A?

2007-05-17 06:19:21 · 3 answers · asked by bridgett l 1 in Science & Mathematics ➔ Mathematics

3 answers

I am reading this as sin A = 12 / 13, find sin 2A?

sin A = 12 / 13
cos A = 5 / 13 (from right angled triangle)
sin 2A = 2 sinA.cosA
sin 2A = 2 x 12 / 13 x 5 / 13
sin 2A = 120 / 169

2007-05-17 09:30:40 · answer #1 · answered by Como 7 · 1⤊ 0⤋

If sin 2A =12/13, then cos 2A = 5/13.
So sin A = sqrt ((1-5/13)/2) = sqrt (4/13)= (2/13)sqrt(13).
A = approximately .588 radians and 2A = approximately 1.17 radians.

2007-05-17 06:38:38 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋

12/13 ?

2007-05-17 06:27:56 · answer #3 · answered by Anonymous · 0⤊ 0⤋