# Thread: help with angle and figure

1. ## help with angle and figure

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

2. This should have been posted in the Trigonometry subforum.

\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

3. 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)?

4. 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
AD = ABcosθ + BCsinθ