# 20-21 Width of the second corridor for a pole to pass horizontally

**Problem 20**

A pole 27 feet long is carried horizontally along a corridor 8 feet wide and into a second corridor at right angles to the first. How wide must the second corridor be?

**Solution 20**

$b = w \csc \theta$

$a + b = 24$

$8 \sec \theta + w \csc \theta = 24$

$w = \dfrac{24 - 8\sec \theta}{\csc \theta}$

$w = 24\dfrac{1}{\csc \theta} - 8\dfrac{\sec \theta}{\csc \theta}$

$w = 24\sin \theta - 8\dfrac{\sin \theta}{\cos \theta}$

$w = 24\sin \theta - 8\tan \theta$

$\dfrac{dw}{\theta} = 24 \cos \theta - 8 \sec^2 \theta = 0$

$24\cos \theta = 8 \sec^2 \theta$

$24\cos \theta = \dfrac{8}{\cos^2 \theta}$

$\cos^3 \theta = \dfrac{8}{24}$

$\cos^3 \theta = \dfrac{1}{3}$

$\theta = 46.10^\circ$

$w = 24 \sin \theta - 8 \tan \theta$

$w = 24 \sin 46.10^\circ - 8 \tan 46.10^\circ$

$w = 8.98 \, \text{ ft}$ *answer*

**Problem 21**

Solve Problem 20 if the pole is of length $L$ and the first corridor is of width $C$.

**Solution 21**

$b = W \csc \theta$

$a + b = L$

$C \sec \theta + W \csc \theta = L$

$W = \dfrac{L - C \sec \theta}{\csc \theta}$

$W = \dfrac{L}{\csc \theta} - \dfrac{C \sec \theta}{\csc \theta}$

$W = L\sin \theta - \dfrac{C \sin \theta}{\cos \theta}$

$W = L\sin \theta - C \tan \theta$

$\dfrac{dW}{d\theta} = L\cos \theta - C \sec^2 \theta$

$L \cos \theta = C \sec^2 \theta$

$L \cos \theta = \dfrac{C}{\cos^2 \theta}$

$\cos^3 \theta = \dfrac{C}{L}$

$\cos \theta = \dfrac{C^{1/3}}{L^{1/3}}$

$W = L \sin \theta - C \tan \theta$

$W = L \left( \dfrac{\sqrt{L^{2/3} - C^{2/3}}}{L^{1/3}} \right) - C \left( \dfrac{\sqrt{L^{2/3} - C^{2/3}}}{C^{1/3}} \right)$

$W = L^{2/3}\sqrt{L^{2/3} - C^{2/3}} - C^{2/3}\sqrt{L^{2/3} - C^{2/3}}$

$W = \left(L^{2/3} - C^{2/3}\right)\sqrt{L^{2/3} - C^{2/3}}$

$W = \left(L^{2/3} - C^{2/3}\right)\left(L^{2/3} - C^{2/3}\right)^{1/2}$

$W = \left(L^{2/3} - C^{2/3}\right)^{3/2}$ *answer*

(You may check the answer of Problem 20 above by using this formula)