[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

Using $\cos\beta = \frac{4}{5}$, we get $\sin\beta = \frac{3}{5}$, so $\sin^2\beta = \frac{9}{25}$. Also,

Substitute:

Thus $f'(\beta) = -\frac{p}{q} = -\frac{96}{25}$, so $p = 96$, $q = 25$, and

[/THOUGHT] $\boxed{721}$