Angle A is in standard position with its terminal arm in quadrant III, the Trigonometry ratio sinA=-3/7 is given, how do you find the values of the other ratios?

There are, broadly speaking, two approaches.

No pictures or thinking about pictures (or not much)

#sin^2A + cos^2 A = 1#.

We are given: #sinA = -3/7#, so we can calculate:

#(-3/7)^2 + cos^2A = 1#

#cos^2A = 40/49#

This tells us that:

#cosA = +- sqrt(40/49) = +- (2sqrt10)/7#

Now the fact that A has terminal side in Q III tells us the #cosA# is negative, so

#cosA = - (2sqrt10)/7#

Now that we have #sinA# and #cosA#, we can get the other values using:

#tanA = sinA/cosA#, #secA = 1/cosA# and so on.

Method 2, thinking about the definitions of the trigonometric ratios:

There are several possible approaches to this:

2a:
Think of #A# corresponding to a point #(x, y)# on the unit circle:
#x^2+y^2=1#

with #sinA = y#, we see immediately that #y=-3/7#

Now use #x^2+y^2=1# to get #x= +- sqrt(40/49) = +- (2sqrt10)/7#

And the fact that we are in Q III, to get #x= - (2sqrt10)/7#

#cosA = x#, #tanA = y/x# and so on.

2b

for points on the terminal side #(x,y)# with #r = sqrt(x^2+y^2)#

and #sinA = y/r = (-3)/7#

We can choose one of these points (the unit circle approaches chooses #r=1#)

Let's make #y= -3#, so we must have #r=7# and, using #r = sqrt(x^2+y^2)#, we get #x=+- 2sqrt10#

In Q III, we must have #x=- 2sqrt10#

Now:

#cosA = x/r#, #tanA=y/r#, and so on.