How do you write the equation #r=5sec(theta-60^circ)# in rectangular form?

Question

1680 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-15T10:45:14

#x+sqrt3 y-10=0.#

graph{((x-2.5)^2+(y-2.5sqrt3)^2-.01)(x+sqrt3y-10)(y-1.73x)=0x^2 [-10, 10, -5, 5]} If O is the pole r =0, A is #(5, 60^o)# then ,

using projection #OA = 5 = OP cos angleAOP=rcos(theta-60^o)#.

# P(r, theta)# moves such that #angleOPA# is #90^o#.

The conversion formula is #r(cos theta, sin theta)=(x, y)#.

Now,

#5=rcos(theta-60^o)#

# = (rcostheta)cos 60^o+(rsin theta)sin 60^o#

#=1/2x+sqrt3/2 y#

This makes intercept 10 on the x-axis.

Look at the dot for the reference point #A (5, 60^o)#, in the graph