Trig (i think) problem help? (viewing angle from porch)

Gyroballer1990

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You are renting property along the water, and you want to sit so as to have as much view of the ocean as possible. The condo has 60 feet of porch frontage and a west wall of 55 feet and an east wall of 70 feet.
a.) Find the Maximum viewing angle, z, of the building (answer in degrees)
b.) How far from the left wall should you sit?
 
You are renting property along the water, and you want to sit so as to have as much view of the ocean as possible. The condo has 60 feet of porch frontage and a west wall of 55 feet and an east wall of 70 feet.
a.) Find the Maximum viewing angle, z, of the building (answer in degrees)
b.) How far from the left wall should you sit?

Gyroballer,

I suspect there is an accompanying diagram with this problem that we cannot see. I will assume that the porch faces north and that the 55 and 70 foot walls are perpendicular to the porch and start at either end of the porch.

What have you tried? What class is this for, and what techniques have you learned?

If you draw an arbitrary point, P, on the porch segment and connect it to the far end points of the walls, you create a couple of triangles. The point on the porch segment, P, breaks the 60 foot porch into two smaller segments of lengths “x” and “60-x”. You can then write equations for the angles at point P using inverse trig functions (inverse tangents) in terms of x.

Your “viewing angle” plus the two other angles at P must total 180 degrees.

If you are familiar with calculus, take the derivative of this equation with respect to x, set it equal to zero, and solve for x.
 
Hello, Gyroballer1990!

I completely agree with wjm11 . . .


You are renting property along the water,
and you want to sit so as to have as much view of the ocean as possible.
The condo has 60 feet of porch frontage and a west wall of 55 feet and an east wall of 70 feet.

a) Find the maximum viewing angle, \(\displaystyle \theta\), of the building (answer in degrees)
b) How far from the left wall should you sit?
Code:
                                    C
                                    *
      A                          *  |
      *                       *     |
      |  *                 *        | 70
   55 |     *           *           |
      |      a *  @  * b            |
      * - - - - - * - - - - - - - - *
      B     x     P       60-x      D

The west wall is: \(\displaystyle AB = 55\)
The east wall is: \(\displaystyle CD = 70\)
The porch is: \(\displaystyle BD = 60\)

Let \(\displaystyle x = BP\), then \(\displaystyle 60-x = PD\)

\(\displaystyle \text{Let }\,\theta\, (@) \,=\,\angle APC\)
\(\displaystyle \text{Let }\,\alpha\, (a) \,=\,\angle APB\)
\(\displaystyle \text{Let }\,\beta\, (b) \,=\,\angle CPD\)

\(\displaystyle \text{We see that: }\:\theta + \alpha + \beta \:=\:180^o\)

\(\displaystyle \text{We wish to maximize }\theta\text{, so we will minimize }\phi \:=\:\alpha + \beta\)


\(\displaystyle \text{In }\Delta ABP\!:\;\tan\alpha = \frac{55}{x}\qquad \text{In }\Delta CDP\!:\;\tan\beta = \frac{70}{60-x}\)

\(\displaystyle \text{We have: }\;\tan\phi \;=\;\tan(\alpha + \beta) \;=\;\frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)

. . \(\displaystyle \text{Then: }\;\tan\phi \;=\;\frac{\frac{55}{x} + \frac{70}{60-x}}{1-\frac{55}{x}\cdot\frac{70}{60-x}} \;=\;-15\cdot\frac{x+220}{x^2-60x+3850}\)


I've set it up . . . YOU can finish it.

 
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