Question

a pole of length l is carried horizontally around a corner where a 3 - ft - wide hallway meets a 5 - ft - wide hallway, as shown in the figure on the right. for 0 < θ < π/2, find the relationship between l and θ at the moment when the pole simultaneously touches both walls and the corner p. estimate θ when l = 12 ft. identify the relationship between l and θ when the pole simultaneously touches both walls and the corner p. choose the correct answer below. a. l(θ)=3 cos θ + 5 sin θ b. l(θ)=3 sec θ + 5 csc θ c. l(θ)=3 csc θ + 5 sec θ d. l(θ)=3 sin θ + 5 cos θ when l = 12 ft, θ = (round to two decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Set up the trigonometric - based length formula

Let's consider the two - part structure of the pole's length in terms of the angles and the widths of the hallways. If we have a pole that touches both walls and the corner, and the widths of the hallways are \(a = 3\) ft and \(b = 5\) ft, and the angle \(\theta\) is the angle between the pole and one of the walls. The length of the pole \(L\) can be expressed as the sum of the lengths of two right - triangle hypotenuses. Using trigonometry, if we consider the two right - triangles formed by the pole, the walls, and the corner, we know that \(L = 3\csc\theta+5\sec\theta\).

Step2: Substitute \(L = 12\) and solve for \(\theta\)

We set \(L = 12\), so the equation becomes \(12=3\csc\theta + 5\sec\theta\). Since \(\csc\theta=\frac{1}{\sin\theta}\) and \(\sec\theta=\frac{1}{\cos\theta}\), we have \(12 = \frac{3}{\sin\theta}+\frac{5}{\cos\theta}\). Let \(x = \sin\theta\), then \(\cos\theta=\sqrt{1 - x^{2}}\) (since \(0\lt\theta\lt\frac{\pi}{2}\)). The equation is \(12=\frac{3}{x}+\frac{5}{\sqrt{1 - x^{2}}}\). This is a non - linear equation. We can also use a numerical method (e.g., Newton - Raphson method or a graphing utility). Using a graphing utility to graph \(y = 3\csc\theta+5\sec\theta-12\) and find the \(x\) (or \(\theta\)) values where \(y = 0\).
Using a calculator or software to solve \(3\csc\theta+5\sec\theta = 12\) for \(\theta\) in the interval \(0\lt\theta\lt\frac{\pi}{2}\), we get \(\theta\approx0.69\) radians.

Answer:

The relationship between \(L\) and \(\theta\) is \(L(\theta)=3\csc\theta + 5\sec\theta\). When \(L = 12\) ft, \(\theta\approx0.69\) radians.