finding the value of theta to the nearest second.

• Jul 6th 2008, 08:19 AM
Dave19
finding the value of theta to the nearest second.
1) sin theta= -0.4545 and theta is in the 4th quadrant

2) sec theta= 2/3 and theta is in 1st quadrant
• Jul 6th 2008, 08:40 AM
Chop Suey
1. sin(theta) = -0.4545 = - 4545/10000 = - 909/2000 (reduce it to lowest terms)
You have to first express it in terms of a fraction.

sin(theta) = y/r

So you got y and you got r. Since theta is in 4th quadrant, x is positive and y is negative. Use the Pythagorean theorem to find x and choose the positive root, then replace in cos(theta) = x/r

x^2 + y^2 = r^2

2. Sec is the reciprocal for cos. So if cos(theta) = x/r, then sec(theta) = r/x. You know how to continue.
• Jul 6th 2008, 08:50 AM
Dave19
chop suey
is that the only way to do it and if it is for the first one is the answer around 1700.
• Jul 6th 2008, 09:00 AM
Chop Suey
Well, that's the way I know. And show me what you have done so far. What answer?
• Jul 6th 2008, 09:04 AM
Dave19
x squared= 3 173 719= 1781.49
ya i think im lost and also what does it mean by nearest second do you have to put it in degrees minutes and seconds