given cosØ = 2/3 and cotØ > 0
so Ø must be in quadrant I
make a sketch of a right -angled triangle with hypotenuse 3 and adjacent 2
then y^2 + 2^2 = 3^2
y = √5
and sin Ø - √5/3
tanØ = √5/2
do the second the same way
(Ø will be in quadrant II)
find sin theta and tan theta if cos theta = 2/3 and cot theta is >0
and
find cos theta and cot theta if sin theta = 1/3 and tan theta is <0
thanks! :)
3 answers
so cos would be root 15/ 4
and cot would be 1/ root 15?
one more question, how do you know it is in the second quadrant?
and cot would be 1/ root 15?
one more question, how do you know it is in the second quadrant?
For the second, draw you triangle in quadrant II
hypotenuse = 3 , opposite = 1
x^2 + 1 = 9
x = ±√8 or ±2√2 , but we are in II, so use x - -2√2
cosØ = -1/3
tanØ = - 1/2√2 ---> cotØ = -2√2
I don't know how you possibly got your answers
you have cosØ = 15/4
That would be impossible since the cosine of any angle lies between -1 and +1
I knew the angle was in the second quadrant from the CAST rule. I hope you have been taught that simple pneumonic , it tells you the sign of the 3 main trig functions in each of the 4 quadrants. Do a simple Google search for "CAST rule".
(the sine is positive in I and II,
the tangent is negative in II and IV , so where is the angle ? )
hypotenuse = 3 , opposite = 1
x^2 + 1 = 9
x = ±√8 or ±2√2 , but we are in II, so use x - -2√2
cosØ = -1/3
tanØ = - 1/2√2 ---> cotØ = -2√2
I don't know how you possibly got your answers
you have cosØ = 15/4
That would be impossible since the cosine of any angle lies between -1 and +1
I knew the angle was in the second quadrant from the CAST rule. I hope you have been taught that simple pneumonic , it tells you the sign of the 3 main trig functions in each of the 4 quadrants. Do a simple Google search for "CAST rule".
(the sine is positive in I and II,
the tangent is negative in II and IV , so where is the angle ? )
2 answers