I would leave the y value in exact form
y = r sinØ
= -14(√3/2) = -7√3
you also missed the negative sign in your y value
exact point (-7, -7√3)
Is this right?
Convert the polar coordinate (-14, 5pi/3) to rectangular form.
Does it mean rectangular coordinates and if so can you check my work?
answer:
r=-14
theta=5pi/3
x=-14 cos 5pi/3
x=-14(0.5)
x=-7
y=rsintheta
y=-14 sin 5pi/3
y=-14(-0.866)
y=12.124
(-7,12.124)
2 answers
actually, you are correct, since 5pi/3 is in QIV, so with r = -14, the point is in QII, with y positive.
2 answers