pls help me solve part(a). thank you.

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- Jun 27th 2009, 08:01 PMhelloyinghelp with angle and figure
pls help me solve part(a). thank you.

- Jun 27th 2009, 08:37 PMyeongil
This should have been posted in the Trigonometry subforum. (Wink)

$\displaystyle \begin{aligned}

\tan(45^{\circ} + \alpha) &= p \\

\frac{\tan 45^{\circ} + \tan \alpha}{1 - \tan 45^{\circ}\tan \alpha} &= p \\

\frac{1 + \tan \alpha}{1 - \tan \alpha} &= p \\

1 + \tan \alpha &= p(1 - \tan \alpha) \\

1 + \tan \alpha &= p - (\tan \alpha)p \\

\end{aligned}$

$\displaystyle \begin{aligned}

(\tan \alpha)p + \tan \alpha&= p - 1 \\

(\tan \alpha)(p + 1) &= p - 1 \\

\tan \alpha &= \frac{p - 1}{p + 1}

\end{aligned}$

What you have scribbled on the top is actually correct, just not simplified:

$\displaystyle \frac{1 - p}{-1 - p} = \frac{-(p - 1)}{-(p + 1)} = \frac{p - 1}{p + 1}$

01 - Jun 28th 2009, 05:23 AMhelloying
Hi actually i meant the 1st part as in part (a). but thanks for telling me the simplied ans for the part(i).do u know how to solve part (a)?

- Jun 28th 2009, 07:19 AMsa-ri-ga-ma