Question

Maximal angle optimization problem... Problem is in next post.

Answer 1

I'm having trouble on the trig part [tan a , tan b, tan theta]. The rest of the problem I can manage to solve.

Answer 2

And I figured out that\[\tan\ \alpha = \frac{2}{1-t},\ \tan \beta= \frac{1}{1+t}\] But how about \[tan\ \theta\ \ ?\]

Answer 3

ok sorry i was away for a bit there
you will need to use sum difference identities for tangent
\[\tan (x \pm y) = \frac{\tan x \pm \tan y}{1\mp \tan x \tan y}\]
note that
\[\theta = \pi - (\alpha + \beta)\]
\[\tan \theta = \tan (\pi - (\alpha + \beta))\]

Answer 4

All right, just one moment...

Answer 5

There should be something easy that I am missing. It's simply not working. Sorry for asking, but can you show me the steps for this part?

Answer 6

sure...it takes 2 steps
step 1
\[\tan(\pi - (a+b)) = \frac{\tan \pi - \tan (a+b)}{1+\tan \pi \tan (a+b)} = -\tan(a+b)\]
note tan(pi) = 0
step 2
\[-\tan(a+b) = -\frac{\tan a + \tan b}{1-\tan a \tan b} = \frac{\tan a + \tan b}{\tan a \tan b -1}\]
note, i distributed the neg on bottom
now plug in ratios for tan(a) and tan(b)
\[\large \frac{\frac{2}{1-t} +\frac{1}{1+t}}{\frac{2}{1-t} *\frac{1}{1+t}-1} = \frac{2(1+t) +(1-t)}{2-(1-t)(1+t)} = \frac{t+3}{t^{2}+1}\]

Answer 7

Oh, I see. Now I can take it from here. Many thanks!

Answer 8

yw