help with angle and figure

helloying · #1 $Old$ June 27th, 2009, 09:01 PM

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

yeongil · #2 $Old$ June 27th, 2009, 09:37 PM

This should have been posted in the Trigonometry subforum.

$\begin{aligned}\tan(45^{\circ} + \alpha) &= p \\\frac{\tan 45^{\circ} + \tan \alpha}{1 - \tan 45^{\circ}\tan \alpha} &= p \\\frac{1 + \tan \alpha}{1 - \tan \alpha} &= p \\1 + \tan \alpha &= p(1 - \tan \alpha) \\1 + \tan \alpha &= p - (\tan \alpha)p \\\end{aligned}$
$\begin{aligned}(\tan \alpha)p + \tan \alpha&= p - 1 \\(\tan \alpha)(p + 1) &= p - 1 \\\tan \alpha &= \frac{p - 1}{p + 1}\end{aligned}$

What you have scribbled on the top is actually correct, just not simplified:
$\frac{1 - p}{-1 - p} = \frac{-(p - 1)}{-(p + 1)} = \frac{p - 1}{p + 1}$

01

helloying · #3 $Old$ June 28th, 2009, 06:23 AM

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)?

sa-ri-ga-ma · #4 $Old$ June 28th, 2009, 08:19 AM

Quote:

Originally Posted by helloying $View Post$

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