Question

two ladders of length 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 h and/or k. assume the ground is horizontal and perpendicular to both walls. (use degrees for any angle measures in the expression. do not include the degree symbol in your answer.)

Explanation:

Step1: Analyze the right - triangle formed by ladders

We have two right - triangles. One ladder of length $k$ makes a $45^{\circ}$ angle with the ground and the other of length $h$ makes a $75^{\circ}$ angle with the ground. The width of the alley is the sum of the horizontal distances from the base of the ladders to the walls.

Step2: Use trigonometric relations

For the ladder of length $k$ making a $45^{\circ}$ angle with the ground, if we consider the right - triangle formed by this ladder, the ground, and the wall, and let the horizontal distance be $x$. Using the cosine function $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, we have $\cos45^{\circ}=\frac{x}{k}$, so $x = k\cos45^{\circ}=\frac{k}{\sqrt{2}}$.

Step3: For the other ladder

For the ladder of length $h$ making a $75^{\circ}$ angle with the ground, let the horizontal distance be $y$. Using the cosine function $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$, we have $\cos75^{\circ}=\frac{y}{h}$, so $y = h\cos75^{\circ}$. Since $\cos75^{\circ}=\cos(45^{\circ}+ 30^{\circ})=\cos45^{\circ}\cos30^{\circ}-\sin45^{\circ}\sin30^{\circ}=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$.

Step4: Calculate the width of the alley

The width $w$ of the alley is $w = k\cos45^{\circ}+h\cos75^{\circ}=\frac{k}{\sqrt{2}}+h\times\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{2k + h(\sqrt{6}-\sqrt{2})}{4}$

Answer:

$\frac{2k + h(\sqrt{6}-\sqrt{2})}{4}$