Draw a right triangle
x=-1 and y=sqroot 3
solve for the hypotenuse by Pythagorean theorom h=2
x is adjacent y is opposite
sin= opp/hyp= Sqroot3/2
cos=adj/hyp=-1/2
tan = opp/adj= sqroot3
Cavan R.
asked 09/17/20I am stuck on this question, i am not sure how to do this, any help would be greatly appreciated!
Draw a right triangle
x=-1 and y=sqroot 3
solve for the hypotenuse by Pythagorean theorom h=2
x is adjacent y is opposite
sin= opp/hyp= Sqroot3/2
cos=adj/hyp=-1/2
tan = opp/adj= sqroot3
Jeffrey K. answered 09/24/20
Together, we build an iron base in mathematics and physics
Cavan, here's how I would answer this:
If you draw the coordinate plane, it's easy to see that (-1, √3) lies in Q2
Drop a perpendicular from the point (-1, √3) to the X-axis. Now we have a right triangle with sides -1 and √3
By Pythagoras (-1)2 + (√3)2 = h2 where h = hypotenuse
1 + 3 = h2
=> h = 2
Let's denote the angle by φ
sin φ = √3/2 sin is +ve in Q2
cos φ = -1/2 cos is -ve in Q2
tan φ = - √3 tan is -ve in Q2
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Cavan R.
thank you :)09/17/20