Question 1083096
Without a picture with side lengths and angle {{{alpha}}} labeled,
or some other way to figure out if angle {{{alpha}}} is opposite or adjacent to the 12-cm leg,
I cannot give you the trigonometric functions of angle {{{alpha}}} ,
but I can draw your triangle as triangle ABC, with right angle at A,
and give you the trigonometric functions of acute angles B and C.
You will have o figure out which one is {{{alpha}}} .
{{{drawing(350,200,-1,13,-1.5,6.5,
triangle(0,0,12,0,0,5),rectangle(0,0,0.5,0.5),
locate(-0.3,0,A),locate(12,0,B),locate(-0.3,5.5,C),
locate(5,0,12cm),locate(4,3,13cm),locate(0.1,3,x)
)}}} B is the angle adjacent to the 12-cm leg, C is opposite the 12-cm leg.
You need the length, {{{x}}} , of leg AC.
For that you use the Pythagorean theorem.
Applied to this triangle, it says
{{{x^2+(12cm)^2=(13cm)^2}}}
{{{x^2+144cm^2=169cm^2}}}
{{{x^2=169cm^2-144cm^2}}}
{{{x^2=25cm^2}}}
{{{x=5cm}}}
With that length you can calculate values of all trigonometric functions for the acute angles.
Then you use the definitions for trigonometric ratios for a right triangle
{{{sin(angle)="opposite_leg"/hypotenuse}}} ,  and {{{cos(angle)="adjacent_leg"/hypotenuse}}} .
You may know tangent defined as
{{{tan(angle)=sin(angle)/cos(angle)}}} ,
or you may have learned it as
{{{tan(angle)="opposite_leg"/"adjacent_leg"}}}  .
If you are  expected to calculate cotangent, secant and cosecant, you would use
{{{cot(angle)=1/tan(angle)}}} , {{{sec(angle)=1/cos(angle)}}} and {{{csc(angle)=1/sin(angle)}}} .
As a tutor, I would prefer to see you figure it out from this point on,
but I will give you the answers because
you are not here to be persuaded to do the work,
and the website tells me that tutors "MUST SHOW ALL WORK."
{{{matrix(3,7,
angle,cosine,sine,tangent,cotangent,secant,cosecant,
B,12/13,5/13,5/12,12/5,13/12,13/5,
C,5/13,12/13,12/5,5/12,13/5,13/12)}}}