# Showing(2)

• January 23rd 2010, 06:25 PM
Punch
Showing(2)
http://i952.photobucket.com/albums/a...g/P1240011.jpg

I have solved for part (a) and part (b), including showing OA, problem lies in showing OM... would appreciate some help for showing OM= $2cos\theta+5sin\theta$
• January 23rd 2010, 07:19 PM
Jhevon
Quote:

Originally Posted by Punch
http://i952.photobucket.com/albums/a...g/P1240011.jpg

I have solved for part (a) and part (b), including showing OA, problem lies in showing OM... would appreciate some help for showing OM= $2cos\theta+5sin\theta$

Using triangle OMA, we have $\sin \theta = \frac {OM}{OA}$. The result follows easily from there.

also, how did you find OA? seems like you did a lot of work judging from the faint lines i see on the diagram.
• January 23rd 2010, 07:34 PM
Punch
Thanks!!!
http://i952.photobucket.com/albums/a...g/P1240011.jpg

For OA, its far more simple! Draw a line perpendicular to the x-axis from the point (5,2). Let point on perpendicular line on x-axis be X.
So
$OX=5$.

$tan\theta= \frac{2}{XA}$

$cot\theta=\frac{XA}{2}$

$XA=2cot\theta$

$OA = 5+2cot\theta$
• January 23rd 2010, 07:48 PM
Jhevon
Quote:

Originally Posted by Punch
Thanks!!!
For OA, its far more simple! Draw a line perpendicular to the x-axis from the point (5,2). Let point on perpendicular line on x-axis be X.
So
$OX=5$.

$tan\theta= \frac{2}{XA}$

$cot\theta=\frac{XA}{2}$

$XA=2cot\theta$

$OA = 5+2cot\theta$

ok. i just observed that $AX = OA - 5$ and jumped to $\cot \theta = \frac {OA - 5}2$ right off the bat. not much different from you. good job