SOLUTION: Please can someone help me solve this problem cot A=(3/4) and its in the 3rd quad. find the remaining ratios sin,cos,tan,csc,sec

Question 28414: Please can someone help me solve this problem
cot A=(3/4) and its in the 3rd quad. find the remaining ratios sin,cos,tan,csc,sec
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
Please can someone help me solve this problem
cot A=(3/4) and its in the 3rd quad. find the remaining ratios sin,cos,tan,csc,sec
Consider cotA = 3/4
Draw a rough sketch of a right angled triangle ABC say with angle C= 90 degrees
If the side opposite to angle A that is if BC= (4x) and the side adjacent to angle A that is AC= (3x), then tan A = 4/3 and cot A = 3/4
A is an angle in the 3rd quadrant and hence here tanA and cotA are positive.
The other four trig ratios are negative.
Using Pythog. theorem, AC^2 + BC^2 = AB^2 (here AB is the hypotenuse)
We get AB, the hypotenuse(the side opposite the right angle) = 5x
Write tanA = 1/cotA and get the other four ratios (numerical) from the rough sketch and attach a minus to each of the four.
Note: If you really understand the concept that the sides ought to be, 3x, 4x and 5x, then you may leave out x in the sense you may assume x=1 without loss of generality.
tanA =4/3
sinA= -(opp/hyp) = -4/5
cosecA = -(hyp/opp) = -5/4
cosA = -(adj/hyp) = -(3/5)
secA = -(hyp/adj) = -(5/3)
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