3779.309 is the answer....
I am pretty sure you can only make one.
If two sides are known, and one of the angles, then the other bits can be deduced and are fixed.
<span>sin can be positive in 1st and 2nd quadrants.
cos can be negative in 2nd and 3rd quadrants.</span>
Answer: Choice B. sqrt(2)
Draw out a right triangle in quadrant IV as you see in the attached image below. The horizontal and vertical legs are both 1 unit long. To ensure that the signs are properly set up, I am making the vertical leg BC have a label "-1" to mean this is below the x axis. Note how
tan(theta) = opposite/adjacent = BC/AB = -1/1 = -1
Use the pythagorean theorem to find that the hypotenuse AC is sqrt(2) units long
a^2 + b^2 = c^2
(1)^2 + (1)^2 = c^2
2 = c^2
c^2 = 2
c = sqrt(2)
The secant of theta is the ratio of the hypotenuse over the adjacent side, so we end up with
sec(theta) = hypotenuse/adjacent
sec(theta) = AC/AB
sec(theta) = sqrt(2)/1
sec(theta) = sqrt(2) which is why choice B is the answer
1 = Hundred tens of thousandths.
0 = tens of thousandths.
6 = thousandths.
5 = hundredths.
3 = tenths.
4 = units.