Real World Problem

**mcinteer** · September 15th 2009, 09:51 AM

Hello,

I work for a small manufacturing firm. We are trying to work out the technical details of how to figure out whether a piece of hardware for a military gun is right or not. This has led me down the road of some algebra that has me stumped.

What I have is:
K/tan(B) = (0.108-K)/tan(A)

All I need to do is to manipulate this down to a "K=" formula, then I can program this into our measurement equipment and away we'll go. But right now I am stuck.

I will be grateful for any help that might be available.

**HallsofIvy** · September 15th 2009, 10:50 AM

Originally Posted by mcinteer

Hello,

I work for a small manufacturing firm. We are trying to work out the technical details of how to figure out whether a piece of hardware for a military gun is right or not. This has led me down the road of some algebra that has me stumped.

What I have is:
K/tan(B) = (0.108-K)/tan(A)

All I need to do is to manipulate this down to a "K=" formula, then I can program this into our measurement equipment and away we'll go. But right now I am stuck.

I will be grateful for any help that might be available.

The basic rule you need is $\frac{a+ b}{c}= \frac{a}{c}+ \frac{b}{c}$ .

$\frac{K}{tan(B)}= \frac{0.108- K}{tan(A)}= \frac{0.108}{tan(A)+ \frac{K}{tan(A)}}$
Subtract $\frac{K}{tan(A)}$ from both sides:
$\frac{K}{tan(B)}- \frac{K}{tan(A)}= K\left(\frac{1}{tan(B)}-\frac{1}{tan(A)}\right)= \frac{0.108}{tan(A)}$
Get common denominators to subtract those fractions:
$\frac{tan(A)}{tan(A)tan(B)- \frac{tan(B)}{tan(A)tan(B)}= \frac{tan(A)- tan(B)}{tan(A)tan(B)}$
so
$K\frac{tan(A)- tan(B)}{tan(A)tan(B)}= \frac{0.108}{tan(A)}$
and divide both sides by that fraction on the left:
$K= \frac{0.108}{tan(A)}\frac{tan(A)tan(B)}{tan(A)- tan(B)}$
$K= \frac{0.108(tan(A)- tan(B)}{tan(B)}$

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