# Cant remember for trig final.

• May 31st 2010, 07:13 AM
acherucheril
Cant remember for trig final.
The length of the normal from a line to the origin is 2 * square root(3). The angle from the positive x-axis to the normal is 60 degrees. What is the equation of the line?

- Probably as simple as plugging numbers into the equation but i dont know what equation...
• May 31st 2010, 07:22 AM
Quote:

Originally Posted by acherucheril
The length of the normal from a line to the origin is 2 * square root(3). The angle from the positive x-axis to the normal is 60 degrees. What is the equation of the line?

- Probably as simple as plugging numbers into the equation but i dont know what equation...

hi

By trigo , you can find the length of one of the sides and the other using phythagoras.The point which the normal cuts the line is also obvious when you know the horizontal and vertical lengths. You can subsequently find the gradient of the normal , then use the fact that m1m2=-1 to get the gradient of the line .
• May 31st 2010, 07:29 AM
skeeter
Quote:

Originally Posted by acherucheril
The length of the normal from a line to the origin is 2 * square root(3). The angle from the positive x-axis to the normal is 60 degrees. What is the equation of the line?

- Probably as simple as plugging numbers into the equation but i dont know what equation...

there probably is a canned formula for this, but I can't remember either.

make a sketch and note that three 30-60-90 triangles are formed by the normal, the x-axis, the y-axis, and the line.

using the known side length ratios, I get the x-intercept at the point $\displaystyle (4\sqrt{3} , 0)$ and the y-intercept as $\displaystyle (0,4)$

coming up with an equation should be rather simple from this point.