SOLUTION: What Would be The solution to these problems? 1.Change -165.852 to degrees,minutes and second's 2.Write 28 42' 13" as a decimal to the nearest thousanth 3.If a 570 angle is in s

Question 172835: What Would be The solution to these problems?
1.Change -165.852 to degrees,minutes and second's
2.Write 28 42' 13" as a decimal to the nearest thousanth
3.If a 570 angle is in standard position,etermine a coterminal angle that is beetween 0 and 360. State the quadrant or axis in which the terminal side lies.
4.Find the value of cosine for angle A in a right triangle ABC if a=3 and b=6.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
1. The degrees part you can see by inspection, it is just the integer part of -165.852. Then we have 0.852 left over. Since there are 60 minutes in a degree, we should hav 0.852*60 degrees. The calculator says: 51.12. Take the integer part again to give you the number of minutes, 51. Now ther is 0.12 left over. There are 60 seconds in a minute, so we have 0.12*60 seconds. Calculator says: 7.2. Putting it all together -165.852� = -165� 51' 7.2"

2. Turn the process around. 28� 42' 13" 13" = (13/60)' or 0.217 (roughly) so that means we have 28� 42.217' But 42.217' = (42.217/60)� or 0.704 (again, roughly) so that means we have 28.704�

3. Use integer division (the kind you first learned to do where you came up with a quotient and a remainder) 570�/360� = 1 r 210. The remainder is the measure of the desired coterminal angle. So 210� is the measure required. Quadrant I is from 0� to 90�, Quadrant II is from 90� to 180�, Quadrant III is from 180� to 270�, and Quadrant IV is what is left. 180� < 210� < 270�, so your angle is in Quadrant III.

4. The first thing you need is the length of the hypotenuse of the triangle given by . Presuming your triangle is labled such that side is adjacent angle A as is typical, . However, common practice dictates that you rationalize any denominators that contain radicals, so:

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