Question 1203785
{{{sin(theta)=4/5}}}, {{{cos(theta)=-3/5}}}
<pre>
You should memorize which trig ratios are positive and which ones are
negative in which quadrants. 

In QI, <font size=6><b>ALL</b></font> 6 trig ratios are positive.
In QII, Only the <font size=6><b>SINE</b></font> and its reciprocal, the cosecant, are positive and the other 4 are negative.
In QIII, Only the <font size=6><b>TANGENT</b></font> and its reciprocal, the cotangent, are positive and the other 4 are negative.
In QIV, Only the <font size=6><b>CALCULUS</b></font> and its reciprocal, the secant, are positive and the other 4 are negative. 

An easy mnemonic way to memorize this is to memorize this sentence for the
basic 3 trig functions sine, cosine, and tangent.

<font size=6><b>A</b></font>LL <font size=6><b>S</b></font>TUDENTS <font size=6><b>T</b></font>AKE <font size=6><b>C</b></font>ALCULUS.

Back to your problem:

{{{sin(theta)=4/5}}}, {{{cos(theta)=-3/5}}}

The sine, 4/5, is a POSITIVE number, so if you've memorized the above, you
know that the sine is only positive in QI and QII
The cosine, -3/5 is a NEGATIVE number, so if you've memorized the above, you
know that the cosine is only positive in QI and QIV, so it is negative only in QII and QIII.

Therefore we know that θ is in QII. So we draw a right triangle in QII,
with one side on the x axis. Angles are always measured counter-clockwise from
the right side of the x-axis. So we draw an arc to indicate the angle θ.
The x value is the adjacent side, the y-value is the opposite side, the r-value is the hypotenuse. 

{{{drawing(400,400,-6,6,-6,6, line(-7,0,7,0), line(0,-7,0,7),
red(line(0,0,-3,0),line(0,0,-3,4),line(-3,4,-3,0),arc(0,0,2,-2,0,127),locate(.2,.75,theta)),green(locate(-2,.75,ref(theta))),
locate(-2.3,0,x=-3),locate(-4.05,2,"y=+4"),locate(-1.3,2.3,"r=+5"))}}} 

Angle θ is OUTSIDE the triangle and the reference angle is INSIDE the
triangle, indicated in green by ref(θ).  We use only positive numbers to
calculate the reference angle.

We find the value of the reference angle by calculator using ONLY POSITIVE
NUMBERS. The sine of the reference angle is its opposite 4, over the
hypotenuse 5. So on the TI-84, in degree mode, we press 

2nd, sin, 4, ÷, 5, 

We see this:

{{{drawing(70,40,0,2,-1,1.7,locate(0,.9,"sin"^(-1)),locate(1.4,1.3,(4/5)) )}}}
           {{{53.13010235}}}

So the reference angle is 53.13010235<sup>o</sup>.

The actual angle is measured by the red arc, so we subtract that value
from 180<sup>o</sup>, and get 126.8698977<sup>o</sup>.
     
126.8698977<sup>o</sup>  <--answer

Edwin</pre>