Question 726188
You use the trigonometric ratios
{{{sin(angle)=opposite_leg/hypotenuse}}}
{{{cos(angle)=adjacent_leg/hypotenuse}}}
 
For most angles, you would let your calculator figure out the values of the sine and cosine of that angle. Fifty to forty years ago, we used tables from books, or slide rules.
 
Sine and cosine of some angles can be calculated more easily.
A right triangle with acute angles measuring {{{30^o}}} and {{{60^o}}} is half of an equilateral triangle. If the length of the short leg (opposite the {{{30^o}}} angle) is {{{a}}} , the length of the hypotenuse will be {{{2a}}}. and according the the Pythagorean theorem, the length of the other leg, {{{b}}}, will be
{{{sqrt((2a)^2-a^2)=sqrt(3a^3)=a*sqrt(3)}}}.
{{{drawing(300,300,-1,11,-6,6,
triangle(0,0,8.66,5,8.66,-5),
line(0,0,8.66,0),rectangle(8.16,0,8.66,0.5),
locate(8.3,2.8,a), locate(8.3,-2.3,a),
locate(4.5,-2,a),locate(4.5,2.7,a),
locate(4.5,0.8,a*sqrt(3)),locate(1,1,30^o),
locate(7.7,4.8,60^o)
)}}}
So, from the point of view of the {{{30^o}}} angle,
{{{sin(30^o)=a/2a=1/2}}} and {{{cos(30^o)=a*sqqrt/2a=sqrt(3)/2}}} and
from the point of view of the {{{60^o}}} angle,
{{{cos(60^o)=a/2a=1/2}}} and {{{sin(60^o)=a*sqqrt/2a=sqrt(3)/2}}}
Ina any case, the legs of a right triangle with acute angles measuring {{{30^o}}} and {{{60^o}}} are as long as
{{{1/2}}} and {{{sqrt(3)/2}}} times the length of the hypotenuse.
 
A right triangle with both acute angles measuring {{{45^o}}} is an isosceles right triangle. Its legs, being opposite angles with equal measure, have equal lengths. Such a triangle is half of a square, and the legs can be proven to be {{{sqrt(2)/2}}} times as long as the hypotenuse.
{{{drawing(300,300,-1,12,-1,12,
rectangle(0,0,10,10),line(0,0,10,10),
rectangle(10,0,9.5,0.5),locate(9.5,5.3,s),
locate(5.2,0.7,s),locate(5.2,5.4,c)
)}}} The Pythagorean theorem says that {{{s^2+s^2=c^2}}}
Then {{{s^2+s^2=c^2}}} --> {{{2s^2=c^2}}} --> {{{4s^2=2s^2}}} --> {{{sqrt(4s^2)=sqrt(2c^2)}}} --> {{{2s=c*sqrt(2)}}} --> {{{s=c*(sqrt(2)/2)}}}