14-15 Ladder reaching the house from the ground outside the wall

$a = 10 \csc \theta$

$b = 8 \sec \theta$
 

$L = a + b$

$L = 10 \csc \theta + 8 \sec \theta$

$\dfrac{dL}{d\theta} = -10 \csc \theta \cot \theta + 8 \sec \theta \tan \theta = 0$

$8 \sec \theta \tan \theta = 10 \csc \theta \cot \theta$

$4 \left( \dfrac{1}{\cos \theta} \right) \left( \dfrac{\sin \theta}{\cos \theta} \right) = 5 \left( \dfrac{1}{\sin \theta} \right) \left( \dfrac{\cos \theta}{\sin \theta} \right)$

$\dfrac{\sin^3 \theta}{\cos^3 \theta} = \dfrac{5}{4}$

$\tan^3 \theta = 5/4$

$\tan \theta = \sqrt[3]{5/4}$

$\theta = 47.13^\circ$
 

$L = 10 \csc 47.13^\circ + 8 \sec 47.13^\circ$

$L = \dfrac{10}{\sin 47.13^\circ} + \dfrac{8}{\cos 47.13^\circ}$

$L = 25.4 \, \text{ ft}$           answer

$x = b \csc \theta$

$y = c \sec \theta$
 

$L = x + y$

$L = b \csc \theta + c \sec \theta$

$\dfrac{dL}{d\theta} = -b \csc \theta \cot \theta + c \sec \theta \tan \theta = 0$

$c \sec \theta \tan \theta = b \csc \theta \cot \theta$

$c \left( \dfrac{1}{\cos \theta} \right) \left( \dfrac{\sin \theta}{\cos \theta} \right) = b \left( \dfrac{1}{\sin \theta} \right) \left( \dfrac{\cos \theta}{\sin \theta} \right)$

$\dfrac{\sin^3 \theta}{\cos^3 \theta} = \dfrac{b}{c}$

$\tan^3 \theta = \dfrac{b}{c}$

$\tan \theta = \dfrac{b^{1/3}}{c^{1/3}}$
 

$L = b \csc \theta + c \sec \theta$

$L = b \left( \dfrac{\sqrt{b^{2/3} + c^{2/3}}}{b^{1/3}} \right) + c \left( \dfrac{\sqrt{b^{2/3} + c^{2/3}}}{b^{1/3}} \right)$

$L = b^{2/3}\sqrt{b^{2/3} + c^{2/3}} + c^{2/3}\sqrt{b^{2/3} + c^{2/3}}$

$L = (b^{2/3} + c^{2/3})\sqrt{b^{2/3} + c^{2/3}}$

$L = (b^{2/3} + c^{2/3})(b^{2/3} + c^{2/3})^{1/2}$

$L = (b^{2/3} + c^{2/3})^{3/2}$           answer