Section 3.3

#1-6. The reference angle is always the closest angle from the terminal side to the x-axis. It will always be between 0 degrees 90 degrees, or 0 radians and π/2 radians. So, for example, consider 250 degrees. That would be in the 3rd quadrant, so we measure the angle between 250 degrees and 180 degrees, getting 250 - 180 = 70 degrees. For another example, consider 5π/7 radians. That would be in the 2nd quadrant, so we measure the angle between 5π/7 and π, getting π - 5π/7 = 7π/7 - 5π/7 = 2π/7. These are sometimes easiest to draw, and then calculate the angle needed to get to the x-axis.

#7-30. First we must find the reference angle involved. The reference angle for all of these will be angles on our special triangles, 30-60-90 and 45-45-90. Once we find the reference angle, we can go to the correct triangle to find the needed trig value (for the reference angle). Then we must adjust the plus or minus sign, depending on the quadrant of our original angle. For example, find sin(210 degrees). Since 210 degrees in is quadrant 3, the reference angle is 210 - 180 = 30 degrees. We can draw a 30-60-90 triangle, and then read off that sin(30 degrees) = 1/2. But since sine is negative in quadrant 3, our final answer is -1/2.

#41. Ignore the quadrant for the moment. Construct a triangle that says sin θ = 3/5, and also find the missing side of that triangle. Now you should be able to find all six trig functions using the triangle. Lastly, adjust the plus or minus sign of each of those trig function values, since we know we are in quadrant 2.

#42-48. Use the same approach as the hint for #41.

#50-53. There is a formula on page 204 of the book. Use that!

#54. We're going to use the formula on page 204 to find the area of the triangle, but we also must use the formula on page 182 to find the area of the sector. Then we can subtract to find the area of the shaded region.

#55a. The picture below is the same as in the textbook, except I've labeled a few more things. First, I've labeled the two segments that actually make up the pipe x & y. Eventually I will need to find x+y. I also labeled the upper angle (between the y & the 6) as theta. That is because that upper angle and the lower theta angle are complements of the same angle, so they must be equal. Finally, I also drew an extra horizontal line across the hall (near the bend), but its length should also be 9.
Use a trig function to find an expression for x, and then another trig function to find an expression for y. Then add those together to find an expression for x+y, the desired length.

#56a. The picture below is the same as in the textbook, except I've drawn and labeled a few more things. First, I've closed off the top of the rain gutter, which then shows a trapezoid. (Our goal is to find the area of that trapezoid.) I've labeled the height of the trapezoid with h, and I labeled the ends of the upper base of the trapezoid with a couple x's. Lastly I labeled where theta fell inside the trapezoid. (Notice the two right triangles.) What we need to do is identify x & h in terms of theta.
Use a trig function to find an expression for x, and another trig function to find an expression for h. Then use those expressions to find an expression for the area of the trapezoid of the rain gutter. (You might need to look up the formula for a trapezoid area.)



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