- #1
basty
- 95
- 0
How do you find the numerical value sin β for the triangle shown on below image?
I can only find
##\tan β = \frac{AB}{BD} = \frac{2x}{3x} = \frac{2}{3} = 0.666666667##
then
##β = \tan^{-1} 0.666666667 = 0.59°##
then
##\sin β = \sin 0.59° = 0.0103##
Is there another method to find the numerical value of sin β?
I can only find
##\tan β = \frac{AB}{BD} = \frac{2x}{3x} = \frac{2}{3} = 0.666666667##
then
##β = \tan^{-1} 0.666666667 = 0.59°##
then
##\sin β = \sin 0.59° = 0.0103##
Is there another method to find the numerical value of sin β?