How do you write the equation $r = 5 sec (θ - 60^{\circ})$ $r=5sec(theta-60^circ)$ in rectangular form?

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Answer 1 · 2017-01-15T10:45:14

$x + \sqrt{3} y - 10 = 0 .$ $x+sqrt3 y-10=0.$

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 $(5, 60^{o})$ $(5, 60^o)$ then ,

using projection $O A = 5 = O P cos ∠ A O P = r cos (θ - 60^{o})$ $OA = 5 = OP cos angleAOP=rcos(theta-60^o)$ .

$P (r, θ)$ $P(r, theta)$ moves such that $∠ O P A$ $angleOPA$ is $90^{o}$ $90^o$ .

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

Now,

$5 = r cos (θ - 60^{o})$ $5=rcos(theta-60^o)$

$= (r cos θ) {cos 60}^{\oplus} (r sin θ) {sin 60}^{o}$ $= (rcostheta)cos 60^o+(rsin theta)sin 60^o$

$= \frac{1}{2} x + \frac{\sqrt{3}}{2} y$ $=1/2x+sqrt3/2 y$

This makes intercept 10 on the x-axis.

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

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