(800+x)tan33=xtan35
800 tan33 + x tan33 = x tan35
x tan35 - x tan33 = 800
x(tan35 - tan33) = 800
x = 800/(tan35-tan33)
aside from the trig values (which are just constants), that's just good old algebra I, right?
How do you solve the equation(800+x)tan33=xtan35?
4 answers
800 tan 33 = x(tan 35-tan 33)
The problem here is not doing the algebra which is easy, but keeping the accuracy which is hard. When you subtract numbers that are close to each other, you lose significant figures in your calculation.
What we need to do is find an accurate way to calculate tan 35 - tan 33
tan 35 - tan 33 = .7002075382 - .6494075932
= .0508 to 3 significant figures
then
800 tan 33 = x(tan 35-tan 33)
is
x = 800 tan 33/.0508
= 10226 or 10,200 to 3 significant figures
The problem here is not doing the algebra which is easy, but keeping the accuracy which is hard. When you subtract numbers that are close to each other, you lose significant figures in your calculation.
What we need to do is find an accurate way to calculate tan 35 - tan 33
tan 35 - tan 33 = .7002075382 - .6494075932
= .0508 to 3 significant figures
then
800 tan 33 = x(tan 35-tan 33)
is
x = 800 tan 33/.0508
= 10226 or 10,200 to 3 significant figures
Wow - a lesson in precision, as well as catching my typo!
Still, if you just typed in
800 tan 33 /(tan 35-tan 33)
I feel sure that the internal 32-bit precision would avoid any serious rounding error in this case.
Still, if you just typed in
800 tan 33 /(tan 35-tan 33)
I feel sure that the internal 32-bit precision would avoid any serious rounding error in this case.
LOL - I bet they were supposed to do it with a slide rule. Otherwise too easy.
5 answers