Asked by Calvin
The measurement of one side of a right triangle is found to be 9.5 inches, and the angle opposite that side is 26°45' with a possible error of 1.5'.
Approximate the percent error in computing the area of the triangle.
Approximate the percent error in computing the area of the triangle.
Answers
Answered by
MathMate
Area 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.
=(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.
Answered by
MathMate
=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."
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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