Relative Error of Right Triangle Area

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Homework Help Overview

The discussion revolves around estimating the relative error of the area of a right triangle using differentials. The triangle's hypotenuse is given, and one of the acute angles is measured with a specific error margin.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the application of differentials to calculate the area and its relative error. There is a focus on whether angle measurements should be in radians or degrees, particularly regarding the differentiation of the sine function.

Discussion Status

Some participants have provided guidance on the necessity of converting angle measurements to radians for accurate calculations. There is an ongoing exploration of the implications of this conversion on the results, with attempts to recalculate the relative error based on this understanding.

Contextual Notes

Participants are discussing the specific angle measurement in degrees and the associated error in minutes of arc, which may affect the calculations. The original poster questions the correctness of their approach based on the unit of measurement used for angles.

Jimbo57
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Homework Statement


Anyone mind checking my answer for me?

The area of a right triangle with hypotenuse H is

A=1/4H^2sin(2x)

x is one of the acute angles. Use differentials to estimate the relative errors of the area if H = 4cm exactly and x is measured to be 30degrees with an error of measurement of 15 minutes of arc.


Homework Equations



A=1/4H^2sin(2x)
dA=1/2H^2cos(2x)dx
dA/A=2cot(60)(0.25)

The Attempt at a Solution


A=1/4*16sin(60)=3.464cm^2
dA=8cos(60)(0.25)
=0.125cm^2
dA/A=0.125/3.464=0.2887=relative error

Now, did I have to convert to radians in step one? 30 to pi/6? Or is this right?
 
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Jimbo57 said:

Homework Statement


Anyone mind checking my answer for me?

The area of a right triangle with hypotenuse H is

A=1/4H^2sin(2x)

x is one of the acute angles. Use differentials to estimate the relative errors of the area if H = 4cm exactly and x is measured to be 30degrees with an error of measurement of 15 minutes of arc.


Homework Equations



A=1/4H^2sin(2x)
dA=1/2H^2cos(2x)dx
dA/A=2cot(60)(0.25)

The Attempt at a Solution


A=1/4*16sin(60)=3.464cm^2
dA=8cos(60)(0.25)
=0.125cm^2
dA/A=0.125/3.464=0.2887=relative error

Now, did I have to convert to radians in step one? 30 to pi/6? Or is this right?

The formula d(sin(2x))=2*cos(2x)*dx is correct only if x is measured in radians. You can call the 2x part either 60 degrees or pi/3 if you adjust your calculator for degrees or radians, but the dx part definitely needs to be in radians.
 
Dick said:
The formula d(sin(2x))=2*cos(2x)*dx is correct only if x is measured in radians. You can call the 2x part either 60 degrees or pi/3 if you adjust your calculator for degrees or radians, but the dx part definitely needs to be in radians.

Thanks Dick. So the steps I took were correct just needed to convert to radians?

How does this look?
15/60 degrees = 0.00436rad=dx
A=1/4*16sin(pi/3)=3.464cm^2
dA=8cos(pi/3)(0.00436)
=0.01744cm^2
dA/A=0.01744/3.464≈0.005=relative error
 
Jimbo57 said:
Thanks Dick. So the steps I took were correct just needed to convert to radians?

How does this look?
15/60 degrees = 0.00436rad=dx
A=1/4*16sin(pi/3)=3.464cm^2
dA=8cos(pi/3)(0.00436)
=0.01744cm^2
dA/A=0.01744/3.464≈0.005=relative error

That sounds much better.
 

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