Calculus
posted 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.

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. 
=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."