Answer:
a) We can clearly see that the radius is r = 5, now let's find the angle.
We can see that it lies between the angles (2/3)*pi and (5/6)*pi
which one is the angle in between these two?
We can calculate it as:
θ = ( (2/3)*pi + (5/6)*pi)/2 = pi*( 4/6 + 5/6)/2 = pi*(9/6)/2 = pi*(9/12) = pi*(3/4)
But we want to write this in the range:
−2π≤θ<0
Knowing that we have a period of 2*pi
our angle will be equivalent to:
pi*(3/4) - 2*pi = pi*( 3/4 - 2) = pi*(3/4 - 8/4) = pi*(-5/4)
Then this point can be represented as:
(5, (-5/4)*pi)
B) Same as before, but this time we have 0≤θ<2π
The first value of θ that we found is in that range, so this will be:
(5, (3/4)*pi)
C) Here we have −5≤r≤0 and −2π ≤ θ< 0.
For point a, we already know that:
θ = (-5/4)*pi
But the radius part looks tricky, right?
In the standard notation, the variable r (radius) is defined as:
r = IrI
This means that we always use r as a positive number, and if in the notation we write a negative number, then when we work with that we just change the sign.
This means that having r = -5 is exactly the same as r = 5.
Then we can write this point as:
(-5, (-5/4)*pi)
