1)
csc Ф=hypotenuse / opposite=15/9=5/3
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Answer: csc Ф=5/3</span>
2)
answer: the side CT is the hypotenuse because it is in front of the right angle.
3)
cotan Ф=adjacent/ opposite=12/9=4/3
answer: cotan Ф=4/3
4)
sin (-360º)=sin (0º+(-1)(360º)=sin 0º=0
Answer: sin (-360)=0
5)
cos (-90º)= cos (90º)=0=0
answer: cos (-90º)=0
There are the combinations that result in a total less than 7 and at least one die showing a 3:
[3, 3] [3,2] [2,1] [1,3] [2,3]
The probability of each of these is 1/6 * 1/6 = 1/36
There is a little ambiguity here about whether or not we should count [3,3] as the problem says "and one die shows a 3." Does this mean that only one die shows a 3 or at least one die shows a 3? Assuming the latter, the total probability is the sum of the individual probabilities:
1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 5/36
Therefore, the required probability is: 5/36
