Question

a pole of length l is carried horizontally around a corner where a 2 - ft - wide hallway meets a 9 - 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. identify the relationship between l and θ when the pole simultaneously touches both walls and the corner p. choices: a. l(θ)=2 sec θ + 9 csc θ, b. l(θ)=2 sin θ + 9 cos θ, c. l(θ)=2 cos θ + 9 sin θ, d. l(θ)=2 csc θ + 9 sec θ. when l = 15 ft, θ = □ (round to two decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Recall trigonometric - related length formula

If we consider the geometric situation of a pole turning around a corner where the widths of the hallways are \(a = 2\) ft and \(b=9\) ft, the length \(L\) of the pole that can be carried around the corner is given by \(L(\theta)=a\csc\theta + b\sec\theta\). Here \(a = 2\) and \(b = 9\), so \(L(\theta)=2\csc\theta+9\sec\theta\).

Step2: Substitute \(L = 15\) into the equation

We have the equation \(15=2\csc\theta + 9\sec\theta\), which can be rewritten as \(15 = \frac{2}{\sin\theta}+\frac{9}{\cos\theta}\). Let \(x=\sin\theta\), then \(\cos\theta=\sqrt{1 - x^{2}}\) (since \(0\lt\theta\lt\frac{\pi}{2}\)), and the equation becomes \(15=\frac{2}{x}+\frac{9}{\sqrt{1 - x^{2}}}\). This is a non - linear equation. Another way is to use a numerical method. We can rewrite the function \(y = 2\csc\theta+9\sec\theta - 15\) and find the root of the function for \(\theta\in(0,\frac{\pi}{2})\). Using a graphing utility or a numerical solver (such as Newton - Raphson method), we solve for \(\theta\).
Using a calculator or software to solve the equation \(2\csc\theta+9\sec\theta=15\) for \(\theta\in(0,\frac{\pi}{2})\):
Let \(y = 2\csc\theta+9\sec\theta - 15\). We can use the built - in equation - solving function of a scientific calculator (e.g., on TI - 84 Plus: \(Y_1=2/\sin(X)+9/\cos(X)-15\), and then use the zero function to find the root in the interval \((0,1.570796)\) (since \(\frac{\pi}{2}\approx1.570796\))
We get \(\theta\approx0.74\) radians.

Answer:

\(0.74\)