# help with angle and figure

• Jun 27th 2009, 08:01 PM
helloying
help with angle and figure
pls help me solve part(a). thank you.
• Jun 27th 2009, 08:37 PM
yeongil
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 AM
helloying
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 AM
sa-ri-ga-ma
Quote:

Originally Posted by helloying
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)?

Using AB and BC you can find AD as