Asked by Angie

Could I solve (800+x)tan33=xtan35 like xtan33=(x/800)tan35?
Answered by Damon
x tan 33 = x tan 35 - 800 tan 33

I do not know how you got what you got, but I just did this problem a few minutes ago and will go find it I hope.
Answered by Damon
NOTE - Steve left a tan 33 out of 800 tan 33

http://www.jiskha.com/display.cgi?id=1419896043
Answered by Reiny
No

you have to expand (800+x)tan33 to get at the x inside the bracket.
so ...
800tan33 + xtan33 = xtan35
now get all the x terms to one side
800tan33 = xtan35 - xtan33
x is a common factor:
800tan33 = x(tan35-tan33)
divide both sides by tan35-tan33
800tan33/(tan35-tan33) = x

with a good scientific calculator you can now this in one sequence of keystrokes:
800
x
tan33
÷
(
tan35
-
tan33
)
=

to get 10226.9023

This answer satisfies the original equation, it does not satisfy your equation.
How did you possibly come up with that ?
Answered by Damon
agree with Reiny. Note the problem here is really not the algebra but the problem of keeping significant figures when subtracting numbers that are close to each other. However, you must start by doing the algebra correctly :)
Answered by Angie
I divided by 800 on both sides to get xtan33=(x/800)tan 35.
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