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A bit of mathamatics

A bit of mathamatics

A bit of mathamatics

(OP)
Is there any method to go for solving equations below
except for using sin[square]a + cos[square]a = 1
any short tricky method
6.58*cos(a) = 3.81 + .058*B
6.58sin(a) = .0056*B
where a is angle in degree B is a numeric value
In early need of answer
thanking you

RE: A bit of mathamatics

Sounds like a college question to me!  If so, I think  that you need to think about it a little longer!

Nigel Waterhouse
n_a_waterhouse@hotmail.com

A licensed aircraft mechanic and graduate engineer. Attended university in England and graduated in 1996. Currenty,living in British Columbia,Canada, working as a design engineer responsible for aircraft mods and STC's.

RE: A bit of mathamatics

Since you have two equations in two unknowns, the solution to the problem can be derived. As they say though, "All roads lead to Rome." There will be no magic solution or special trick to get you to the answer more quickly than the method of solution you allude to.

Good luck.

RE: A bit of mathamatics

I agree with the "no magic" way. It looks the best way to use sin^2+cos^2=1. This leaves you with an equation with only B unknown.

RE: A bit of mathamatics

I agree with the last two responses.  There are 2 unknowns - a and B - and two equations.  If you don't use the known relationship between cos and sin, then it becomes 3 unknowns, cos(a), sin(a) and B and three unknowns can't be solved from 2 equations.  So you have to use the relationship between cos(a) and sin(a) which you have stated.

RE: A bit of mathamatics

There are a few ways to attack this problem:
a. Pencil and paper with some numerical method for the solution of a system of nonlinear transcendental equations. Modern software often use iteration methods since the computer hardware processes them easily.
b. Graphical methods are good but not much accurate. Incidentally, what kind of accuracy do you need for that solution?
c. The easiest way nowadays is the computer math software. I suggest that you type at your search engine "math software" and some reasonably good packages will appear on the screen. This is what I did long time ago when such packages were inexpensive.
d. I used that software and I got the following results:
a=0.04058 in radians
b=47.66524 in per unit
in the interval a<=3 radians and b>0 in per unit

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