Finding the Radius (word problem)

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Homework Help Overview

The problem involves finding the radius of an arc situated between two blocks of different heights, specifically 1 inch and 2 inches, with the arc measuring 6 inches in length. The context suggests a geometric and trigonometric approach to the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss labeling points and angles related to the arc and blocks, suggesting the use of trigonometric functions to derive equations involving the radius and angles. Some express confusion about the equations and the geometric setup, while others explore the possibility of creating right triangles to aid in understanding.

Discussion Status

The discussion is ongoing, with various participants offering hints and suggestions for approaching the problem. Some participants have expressed uncertainty about the equations and their derivations, while others have noted that the problem may not have an elementary solution, indicating a need for numerical methods.

Contextual Notes

Participants have mentioned that the solution may only be an approximation and that there are constraints related to the geometric setup that are still being clarified. There is also a suggestion that the problem has generated significant interest, leading to further exploration of similar problems.

  • #31
Miike012 said:
But yes that is the answer in the back of my trig book... How did you get that?
And I have a TI-89 calc..

I used Maple. With a program like that, there is little work to do. Here's the steps you type to solve it with Maple. I simplified it slightly for clarity:

> equation1 := r*cos(alpha)+1 = r;
> equation2 := r*cos(beta)+2 = r;
> equation3 := r*(alpha+beta) = 6;

> fsolve({equation1,equation2,equation3}, {r, alpha, beta});

{r = 2.768080732, alpha = 0.8779382494, beta = 1.289628589}

> alphadegrees := evalf(180*alpha/Pi); 50.30215635

> betadegrees := evalf(180*beta/Pi); 73.89027526

Takes all the pain and suffering out of it, eh?
 
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  • #32
I have another problem...

Six Identical pipes, each with radius of 1 foot are ties tightly together with a metal band... Find the length of the metal band...

I posted a picture...

My strategy is dividing the figure in half
Then finding the measure of the two arcs in red ( which should be equal)
Then finding the measure of the yellow line..
Once I find that... I can multiply it by three...
I am just not sure of how to do it...
 

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  • #33
erm :redface: … page 3 ? …

for a new problem, best to start a new thread :smile:
 
  • #34
okay
 

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