Question 1138457
1. The largest angle of the triangle whose sides are of length 2cm, 4cm, 
   5cm is the one opposite the longest (5-cm) side.
To find that angle we have to use law of cosines, which says that in a triangle with sides a, b, and c, and angle A opposite side a,
{{{a^2=b^2+c^2-2bc*cos(A)}}}
If A is a right angle, then {{{cos(A)=0}}} , and that equation turns into the {{{a^2=b^2+c^2}}} (the Pythagorean theorem). Otherwise, as in this case, the {{{-2bc*cos(A)}}} "correction" makes a larger for obtuse angles (as in this case), and smaller for acute ones.
So, for this case we find {{{A}}} opposite {{{a=5}}}(cm) , and the other two side lengths are b and c.
{{{5^2=2^2+4^2-2*2*4*cos(A)}}}
{{{25=4+16-16*cos(A)}}}
{{{25=20-16cos(A)}}}
{{{25-20=-16cos(A)}}}
{{{5=-16cos(A)}}}
{{{-5/16=cos(A)}}}
{{{highlight(A=108.2^o)}}}
 
2. If {{{sin(x) =3/5}}} and x is acute, x is the smallest angle of a right triangle with sides measuring 3, 4, and 5:
{{{drawing(250,200,-.5,4.5,-.5,3.5,
triangle(0,0,4,0,4,3),locate(4.1,1.8,3),
locate(1.9,0,4),locate(2,1.5,5),
red(arc(0,0,2,2,-36.9,0)),locate(1,0.5,red(x))
)}}} So, {{{cos(x)=4/5}}} and {{{tan(x)=3/4}}} .
From there, I can find, in a list of trigonometric identities,
{{{tan(2x)=2tan(x)/(1-tan^2(x))}}} and {{{tan(x/2)=sin(x)/(1+cos(x))}}} ,
and calculate
{{{tan(2x)=2(3/4)/(1-(3/4)^2)=(3/2)/(1-9/16)=(3/2)/(7/16)=(3/2)(16/7)=highlight(24/7)}}} , and
{{{tan(x/2)=(3/5)/(1+(4/5))=(3/5)/(9/5)=(3/5)(5/9)=highlight(1/3)}}} 
 
3. Find all the angles between 0° and 360° whose sine is +.5.
We know that {{{sin(30^o)=0.5}}} , and we know that in each quadrant
The function sine takes values from 0 to 1, or 0 to -1.
In the whole first counterclockwise turn, between 0° and 360°,
there is only one angle whose sine is 1 {{{(90^o)}}} (and only one angle whose sign is -1), but all other positive values happen twice,
once as sine goes from 0 to 1 in the first quadrant, and again as sine goes from 1 back to 0 in the second quadrant.
Suplementary angles {{{30^o}}} and {{{180^o-30^o=150^o}}} have the same 0.5 sine.
{{{drawing(300,175,-1.2,1.2,-0.2,1.2,grid(0),
red(circle(0,0,1)),triangle(0,0,0.866,0,0.866,0.5),
triangle(0,0,-0.866,0,-0.866,0.5),arc(0,0,0.4,0.4,-30,0),
arc(0,0,0.4,0.4,-180,-150),green(arc(0,0,1.2,1.2,-150,0)),
locate(0.05,0.55,green(150^o)),locate(0.2,0.2,30^o),
green(triangle(-0.52,0.3,-0.51,0.3,-0.52,0.31)),locate(-0.38,0.2,30^o)
)}}}