Logs and Special Lines.

Let $x(t) = a(\ln(t))^2 + b,\:\ y(t) = c(\ln(t))^3 + d \implies \dfrac{dy}{dx}|(t = t_{1}) = \dfrac{3c}{2a}\ln(t_{1}) \implies$ the tangent line to the curve at $(x(t_{1}),y(t_{1}))$ is: $y - (c(\ln(t_{1}))^3 + d) = \dfrac{3c}{2a}\ln(t_{1})(x - (a(\ln(t_{1}))^2+ b))$

Let the line be normal to the curve at $(x(t_{2}),y(t_{2})) \implies c\ln(\dfrac{t_{2}}{t_{1}})((\ln(t_{2}))^2 + \ln(t_{1})\ln(t_{2}) + (\ln(t_{1}))^2) = \dfrac{3c}{2}\ln(t_{1})(\ln(t_{2}) + \ln(t_{1}))\ln(\dfrac{t_{2}}{t_{1}}) \implies$

$\dfrac{c}{2}\ln(\dfrac{t_{2}}{t_{1}})(2(\ln(t_{2}))^2 - \ln(t_{1})\ln(t_{2}) - (\ln(t_{1}))^2) = 0$ $t_{1} \neq t_{2} \implies \ln(t_{2}) = -\dfrac{\ln(t_{1})}{2} \implies t_{2} = \dfrac{1}{\sqrt{t_1}}$

Since the tangent is also normal to the curve at $(x(t_{2}),y(t_{2})) \implies \dfrac{9c^2}{4a^2}\ln(t_{1})\ln(t_{2}) = -1$ $\implies \dfrac{9c^2}{8a^2}(\ln(t_{1}))^2 = 1 \implies \ln(t_{1}) = \pm\dfrac{2\sqrt{2}a}{3c} \implies$ the two slopes are $\pm\sqrt{2}$ .

$\tan(\theta) = \sqrt{2} \implies \theta = \arctan(\sqrt{2}) \approx 54.73561 \implies \lambda = 2\theta \approx \boxed{109.47122}$ .