What is the Cartesian form of #(99,(17pi )/12)#?

Question

1380 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-19T18:10:24

#(-95.63, -25.62)#

polar coordinates #(r, theta) = (99, (17pi)/12)# is in quadrant IV

rectangular coordinates (#rsin theta, r cos theta)#

Make sure your calculator MODE is in Radians:

#99 sin (17pi/12) ~~-95.6266568#

#99 cos (17pi/12) ~~ -25.62308547#

Rectangular coordinates: #(-95.63, -25.62)#

Answer 2 · 2017-03-19T18:33:46

#-(99 (sqrt3 -1))/(2sqrt2),-(99 (sqrt3 +1))/(2sqrt2)#

In polar coordinates it is #r=99 and theta=(17pi)/12#. In cartesean form it would be #x= rcos theta# and #y = r sin theta #

Accordingly cartesean coordinates would be #x=99 cos ((17pi)/12)=99 cos(pi +(5pi)/12)= -99 cos((5pi)/12)= -99 cos 75^o=-(99 (sqrt3 -1))/(2sqrt2)#

similarly,#y= 99 sin ((17pi)/12)= -99 sin((5pi)/12)= -99 sin 75^o= -(99(sqrt3 +1))/(2sqrt2)#