Suppose sin 2A = √51/10, and 2A is an angle in the Second Quadrant. (Assume 0 < 2 A < 360 ) Then sin A =

Answer 1

sin2A = sqr51/10, cos2A = sqr(hypotenuse squared - adjacent side squared)

cos2A = -sqr(1-51/100) = -sqr(49/100) = -.7

use the half angle formula

sinA = sqr([1- cos2A)/2 ] = sqr(1--0.7)/2 = sqr(1.7/2) = sqr(0.85) = about 0.922

sinA = 0.922

or

sin2A =sqr51/10 = about 0.714143

2A = about 180-45.573 = 134.427 degrees

A= 134.427/2 = 67.214

sin67.214 = 0.922

sinA = 0.922 rounded off to 3 decimal places

Answer 2

cos2A = - √(1 - 51/100) = - 7/10

sinA = √(1 - cos2A)/2 = √(1 + 7/10)/2 = √17/20 = √85/10