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July 31, 2014

July 31, 2014

Posted by **Calvin** on Wednesday, October 28, 2009 at 11:40pm.

Approximate the percent error in computing the area of the triangle.

- Calculus -
**MathMate**, Thursday, October 29, 2009 at 8:55amArea of right triangle

=(1/2)L*(L)cot(θ)

=(1/2)L²cot(θ)

L=9.5,

θ=26°45' ± 1.5'

=0.46688 rad ± 0.000002315 rad.

If you have not done calculus yet, calculate the area based on the given θ, then calculate the largest and smallest possible value of θ to give the higher and lower limits.

If you have done calculus, set

A(θ)=L²cot(θ) and differentiate with respect to θ to get e=A'(θ). Multiply e by the error in θ to give the error in area.

I get ±0.000516 sq.in. using both methods.

- Calculus -
**MathMate**, Thursday, October 29, 2009 at 9:03am=0.46688 rad ± 0.000002315 rad.

should read 0.46688 rad ±0.000436 rad.

The calculated error should therefore read:

"I get ±0.097 sq.in. using both methods."

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