Trigonometric Form Question

Find the trigonometric form: z = (5/2)((sqrt 3) -i))

Ok. I did the following:
z=r(cosθ + isinθ)
r=12.5
tanθ = 11pi/6 (because it is in quadrant 4)
To end with : z = 12.5(cos(11pi/6) + sin(11pi/6))

However, my final solution is incorrect because i cant match it up to original solution. Where am i going wrong? Any help is appreciated.

Distribute 5/2 so z = 5√3/2 - 5i/2. That is in the algebraic form a + bi.

We want to convert it to the polar form r(cosθ + isinθ) or rcisθ.

To get r, we solve for the square root of the sum of the squares of a and b.

Our a = 5√3/2 and b = -5/2. After solving, we arrive at r = 5.

To get θ, we solve for arctan(b/a). After solving, we arrive at θ = 11π/6.

So the polar form is 5(cos11π/6 + isin11π/6) or simply 5cis11π/6.

r is not equal to 12.5 it is equal to 5 coz

r^2 = [ ((5/2)(sqrt 3))^2 + (5/2)^ 2 ] ^ (1/2)
r^2 = 25^ (1/2)
r = 5.

tan θ = -1 / sqrt 3
θ = -30 degree
z = 5 {cos (-30) + isin (-30 )}
z = 5 {cos 30 - isin 30 }
z = 5 {cos pi /6 - i sin pi /6 }