Problem involving trig functions

In summary, the problem involves finding the optimal distance from a window of known height and position to maximize the visual angle. The solution involves using trigonometric identities and taking derivatives to obtain an equation with only two variables, x and theta. The optimal value can be determined by finding the global turning point of the equation.
  • #1
joess
16
0
[SOLVED] problem involving trig functions

Homework Statement


Your room has a window whose height is 1.5 meters. The bottom edge of the window is 10 cm above your eye level. (See figure) How far away from the window should you stand to get the best view? (“Best view” means the largest visual angle, i.e. angle between the lines of sight to the bottom and to the top of the window.)

6kk3i84.gif


Homework Equations



The Attempt at a Solution


So the question's basically saying that I should maximize theta right? What I have so far is just
tan(O+a) = 1.6/x
and
a = arctan(0.1/x)
[by the way, O=theta, i couldn't figure out how to insert the proper symbol]
But even if I substitute a into the first equation, I still have 2 variables, and I can't figure out how to get only theta in an equation. Am I even on the right track?
 
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  • #2
have you tried using
[tex]\tan (\theta+\alpha) = \frac{\tan \theta + \tan \alpha}{1-\tan \theta \,\tan \alpha}[/tex]

and get an equation involving only x and tan theta, then to maximise, try determine global turning point (by looking at the derivaties)
 
  • #3
But when I take the derivative of tan (O+0.1/x) = 1.6/x, or i change it using that tan identity, I get an equation that involves theta, the derivative of theta, and x, and I'm not sure how to optimize that
 
  • #4
This is what i get when i take the derivative after using the tan identity:

[tex]\frac{x^{2}(sec^{2}\theta\frac{d\theta}{dx} - \frac{0.1}{x^{2}})}{-0.1(xsec^{2}\theta\frac{d\theta}{dx} - tan\theta)} = \frac{-1.6}{x^{2}}[/tex]

I still have both [tex]\theta[/tex] and [tex]x[/tex], so how do I optimize it?

Edit: never mind, I got it. :)
 
Last edited:

1. What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, and tangent. These functions are used to relate the angles and sides of a right triangle.

2. How can I solve a problem involving trigonometric functions?

To solve a problem involving trigonometric functions, you will need to use the given information to set up a trigonometric equation. Then, you can use algebraic techniques to solve for the unknown variable.

3. Can I use trigonometric functions to solve non-right triangles?

Yes, you can use trigonometric functions to solve non-right triangles. However, you will need to use additional formulas and techniques such as the Law of Sines or Law of Cosines.

4. How do I know which trigonometric function to use?

The trigonometric function you use will depend on the given information in the problem. If you are given the measure of an angle and the length of a side, you can use the sine, cosine, or tangent function. If you are given the lengths of two sides, you can use the inverse trigonometric functions (arcsine, arccosine, arctangent) to find the measure of an angle.

5. Can I use trigonometric functions in real-life applications?

Yes, trigonometric functions are used in many real-life applications such as engineering, architecture, physics, and navigation. They can be used to solve problems involving angles and distances, such as determining the height of a building or the angle of elevation for a satellite dish.

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