[THOUGHT] I will derive a formula for the area (f(\theta)) symbolically in terms of (\theta), then differentiate it using Sympy and substitute (\cos\beta = 4/5), (\sin\beta = 3/5) to find (f'(\beta)). From that result, I will extract (p) and (q) and compute (p + q^2). [/THOUGHT] [PYTHON] import sympy as sp

Define symbols

theta = sp.symbols('theta', real=True) c = sp.cos(theta) s = sp.sin(theta)

Define t as function of theta using derived formula t = 4 / (1 + 5c)

t = 4 / (1 + 5*c)

Define f(theta) = area of triangle BCP = 30 * s * (1 - t)

f = 30 * s * (1 - t) f_simplified = sp.simplify(f) print("f(theta) =", f_simplified)

Differentiate f with respect to theta

f_prime = sp.simplify(sp.diff(f_simplified, theta)) print("f'(theta) =", f_prime)

Now substitute cos(beta)=4/5, sin(beta)=3/5

beta_val = sp.acos(sp.Rational(4,5)) # not strictly needed, but for clarity subs_dict = {c: sp.Rational(4,5), s: sp.Rational(3,5)}

Since f_prime is expressed in terms of sin and cos of theta,

we replace sp.cos(theta) and sp.sin(theta) with 4/5 and 3/5.

f_prime_beta = sp.simplify(f_prime.subs({ sp.cos(theta): sp.Rational(4,5), sp.sin(theta): sp.Rational(3,5) })) print("f'(beta) =", f_prime_beta)

Express f'(beta) as -p/q

f_prime_beta_rational = sp.nsimplify(f_prime_beta) print("f'(beta) as rational =", f_prime_beta_rational)

Extract p and q from -p/q

Ensure it's in the form -p/q

num, den = sp.fraction(f_prime_beta_rational) num_simpl = sp.simplify(num) den_simpl = sp.simplify(den) print("numerator:", num_simpl, "denominator:", den_simpl)

We want f'(beta) = -p/q with p,q > 0 and coprime

p = -num_simpl q = den_simpl p = sp.simplify(p) q = sp.simplify(q) print("p =", p, "q =", q)

Compute p + q^2

result = p + q**2 print("p + q^2 =", result) [/PYTHON]

[THOUGHT] From the Python output, we have [ f'(\beta) = -12\sin^2\beta + \frac{15}{2}\cos 3\beta + \frac{78}{25}. ] Using (\cos\beta = \frac{4}{5}), we get (\sin\beta = \frac{3}{5}), so (\sin^2\beta = \frac{9}{25}). Also, [ \cos 3\beta = 4\cos^3\beta - 3\cos\beta = 4\left(\frac{4}{5}\right)^3 - 3\left(\frac{4}{5}\right) = \frac{256}{125} - \frac{300}{125} = -\frac{44}{125}. ] Substitute: [ f'(\beta) = -12\cdot\frac{9}{25} + \frac{15}{2}\left(-\frac{44}{125}\right) + \frac{78}{25} = -\frac{108}{25} - \frac{66}{25} + \frac{78}{25} = -\frac{96}{25}. ] Thus (f'(\beta) = -\frac{p}{q} = -\frac{96}{25}), so (p = 96), (q = 25), and [ p + q^2 = 96 + 25^2 = 96 + 625 = 721. ] [/THOUGHT] \boxed{721}