Question

Find the terminal point determined by t= 10pi/3.

Answer 1

We have that
angle t=$\frac{10 \pi }{3}$

we know that
$\frac{10 \pi }{3}=(\frac{9 \pi }{3}+ \frac{ \pi }{3} )$

the terminal point belong to the III quadrant
so
the x-coordinate is negative
the y-coordinate is negative

Find the x-coordinate  in the unit circle
$x=r*cos t$
$r=1 \\ t= \frac{ \pi}{3} \\ x=-1*cos \frac{ \pi }{3} \\ x=- \frac{1}{2}$

Find the y-coordinate  in the unit circle
$y=r*sin t$
$r=1 \\ t= \frac{ \pi}{3} \\ y=-1*sin \frac{ \pi }{3} \\ y=- \frac{ \sqrt{3}}{2}$

the terminal point is $(- \frac{1}{2},- \frac{ \sqrt{3}}{2} )$

therefore

the answer is
$(- \frac{1}{2},- \frac{ \sqrt{3}}{2} )$

Answer 2

Note: On the unit circle, P (x, y) = (cos(t), sint(t))
Terminal point of t = 10pi/3 is ( -1/2, -sqrt(3)/2)