If Dave is standing next to a silo of cross-sectional radius r=9 feet at the indicated position, his vision will be partially obstructed. Find the portion of the y-axis that Dave cannot see. (Hint: Let a be the x-coordinate of the point where line of sight #1 is tangent to the silo; compute the slope of the line using two points (the tangent point and (12, 0)). On the other hand, compute the slope of line of sight #1 by noting it is perpendicular to a radial line through the tangency point. Set these two calculations of the slope equal and solve for a. Enter your answer using interval notation. Round your answer to three decimal places.)
So I tried to follow the Hint and solve it using that method, but to no avail. Therefore, just to see what the correct answer was, in order to determine what I needed to obtain using the other method, I solved it using the trigonometric method below:
$$t=acos(\frac{9}{12}) \\$$ $$cos(t)= \frac{3}{4} \\$$ $$sin(t)=1-(\frac{3}{4})^2= \frac{\sqrt{5}}{4}$$ $$tan(t)= \frac{\sqrt{5}}{3} \\$$ $$[-12tan(t), 12tan(t)]= [-4\sqrt{5}, 4\sqrt{5}]$$
However, this answer was still marked as incorrect, and the system doesn't say what the correct answer should be.
Did I do something wrong, or is there an error in the system?
Thank you in advance for any help that anyone can provide!
First, an error in your calculation: $$\sin^2(t)=1-\left(\frac{3}{4}\right)^2=\frac{7}{16}\\ \sin(t)=\frac{\sqrt{7}}{4}\\ \tan(t)=\frac{\sqrt{7}}{3}$$ Second, you are looking for $y$ such that $\tan(t)=\frac{12}{y}$ $$y=\frac{12}{\tan(t)}=\frac{36}{\sqrt{7}}=\frac{36\sqrt{7}}{7}\approx 13.607$$