Question

two ladders of length a lean against opposite walls of an alley with their feet touching, as shown in the figure on the right. one ladder extends h feet up the wall and makes a 75° angle with the ground. the other ladder extends k feet up the opposite wall and makes a 45° angle with the ground. find the width of the alley in terms of a, h, and/or k. assume the ground is horizontal and perpendicular to both walls. the width of the alley is given by . (use degrees for any angle measures in the expression. do not include the degree symbol)

Explanation:

Step1: Analyze the right - triangle formed by the 45 - degree ladder

For the ladder that makes a 45 - degree angle with the ground, if we consider the right - triangle formed by the ladder, the ground, and the wall. Let the height it reaches on the wall be \(h\) feet and the distance from the base of the ladder to the wall be \(x\) feet. Since \(\tan45^{\circ}=1=\frac{h}{x}\), we have \(x = h\).

Step2: Analyze the right - triangle formed by the 75 - degree ladder

For the ladder that makes a 75 - degree angle with the ground, if the distance from the base of the ladder to the wall is \(y\) feet and the height it reaches on the wall is \(k\) feet. We know that \(\tan75^{\circ}=\tan(45^{\circ}+ 30^{\circ})=\frac{\tan45^{\circ}+\tan30^{\circ}}{1 - \tan45^{\circ}\tan30^{\circ}}=\frac{1+\frac{\sqrt{3}}{3}}{1 - 1\times\frac{\sqrt{3}}{3}}=\frac{3 + \sqrt{3}}{3-\sqrt{3}}=\frac{(3 + \sqrt{3})^2}{(3-\sqrt{3})(3 + \sqrt{3})}=\frac{9 + 6\sqrt{3}+3}{9 - 3}=2+\sqrt{3}\). And \(\tan75^{\circ}=\frac{k}{y}\), so \(y=\frac{k}{2 + \sqrt{3}}=(2-\sqrt{3})k\).

Step3: Calculate the width of the alley

The width of the alley \(w=x + y\). Substituting \(x = h\) and \(y=(2-\sqrt{3})k\) into the formula, we get \(w=h+(2 - \sqrt{3})k\).

Answer:

\(h+(2 - \sqrt{3})k\)