Question 1073744
<pre><b>
Since we are given 180° < A < 270°, we know A is in QIII, so we
draw angle A in QIII, and since {{{ SINE=OPPOSITE/HYPOTENUSE=y/r}}},
we can put angle A in a right triangle in QIII with OPPOSITE=y as 
the numerator -5 and the denominator as the HYPOTENUSE=r as 7.  
(We always take the hypotenuse r as positive.

Then we find the adjacent side, x, by the Pythagorean theorem:

{{{r^2=x^2+y^2}}}
{{{7^2=x^2+(-5)^2}}}
{{{49=x^2+25}}}
{{{24=x^2}}}
{{{"" +- sqrt(24)=x}}}
{{{"" +- sqrt(4*6)=x}}} 
{{{"" +- 2sqrt(6)=x}}}


Now you can use your formula
We know to take the negative sign since
x goes to the left.

{{{-2sqrt(6)=x}}}

{{{drawing(400,400,-7,2,-7,2,triangle(0,0,-sqrt(24),0,-sqrt(24),-5),
line(-10,0,10,0),line(0,-10,0,10),locate(-6,-2.5,y=-5),locate(-2.5,-2.5,r=7),
locate(-3.5,.5,x=-2sqrt(6)), red(arc(0,0,2,-2,0,226),locate(-1,1,A))))}}} 

Now use your formulas for 2A using OPPOSITE, ADJACENT and HYPOTENUSE from
the triangle graph above.  Be sure to use the correct signs.  The
ADJACENT or x in QIII is negative (because it goes left) and the OPPOSITE 
or y in QIII is also negative (because it goes down). The hypotenuse r is 
ALWAYS positive in every quadrant.

If you still get the wrong answers, tell me in the thank-you note form
below and I'll get back to you by email.  No charge ever! I do this for fun.

Edwin</pre><b>