How do you graph #r=2csc(theta+60)#?

Question

2070 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-09T13:36:19

A straight line that makes intercepts #4/sqrt3 and 4# on #theta=0 and theta=90^o#

This is the polar equation of the straight line #x/(4/sqrt3)+y/4=1# that

makes intercepts #4/sqrt3 and 4# on^theta=0 and theta=pi/2..

Using cos theta =x/r and sin theta = y/r,

rsin(theta+60^o)#

#=r(sintheta cos 60^o +cos theta sin 60^0)#

#=y/2+sqrt3/2x=2# and this, in the intercept form, is

#x/(4/sqrt3)+y/4=1#

graph{y+sqrt3 x=4 [-10, 10, -5, 5]}