Height of a sign when you know the maximum viewing distance is 35ft and the bottom of the sign must be 20 ft off the ground

[matter2003]

So I am having an issue with this because the asnwer I am getting makes no sense so it must be wrong inmy mind.

Here is the question: Suppose I’m hanging a sign over the road and the bottom must be 20 feet off the ground. I want the location of the maximum viewing angle to be 35 feet away. How tall should I make the sign?

So from what I understand(which admittedly might be wrong) I have x, which is the distance away of the maximum viewing angle and is 35 feet. I want to solve for arctan(alpha) = 20+y/35 or am I wanting the derivative of this? If I get the derivative I end up with 35/y^2+40y+1625...am I simply solving for y at that point?

 

[MarkFL]

I'm assuming you are using \(y\) as the height of the sign. I would begin by stating:

[MATH]\tan(\alpha+\beta)=\frac{20+y}{35}[/MATH]
Where:

[MATH]\tan(\alpha)=\frac{20}{35}=\frac{4}{7}[/MATH] and \(\beta\) is the actual viewing angle.

Now, apply the sum of angles identity for tangent function to the first equation...what do you get?

 

[matter2003]

I think I am wanting something like the derivative of arctan(20+y/35) - arctan(20/35) am I not?

 

[MarkFL]

I think I am wanting something like arctan(20+y/35) - arctan(20/35) am I not?

Let's see where my suggestion leads us:

[MATH]\frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}=\frac{20+y}{35}[/MATH]
Now, plug in for \(\tan(\alpha)\):

[MATH]\frac{\dfrac{4}{7}+\tan(\beta)}{1-\dfrac{4}{7}\tan(\beta)}=\frac{20+y}{35}[/MATH]
[MATH]\frac{4+7\tan(\beta)}{7-4\tan(\beta)}=\frac{20+y}{35}[/MATH]
Can you solve for \(\tan(\beta)\) ?