Question 1210628
<pre>

You must be beginning trig, so I'll go a little easier.

{{{sin(p)}}}{{{""=""}}}{{{"opposite"/"hypotenuse"}}}{{{""=""}}}{{{a/c}}}{{{""=""}}}{{{3/5}}}

We can use the numerator 3 for "a", and the denominator 5 for "c".

Sketch a right triangle with angle "p", side opposite angle "p" as "a",
hypotenuse as "c", and side adjacent to angle "p" as "b". 

{{{drawing(200,500/3,0,6,0,5,
rectangle(4.7,1,5,1.3),
line(1,1,5,1), line(5,1,5,4), line(1,1,5,4),

locate(5.1,2.5,a=3),locate(2,2.9,c=5),locate(1.6,1.53,p), locate(3,1,"b=?")

 ) }}}

{{{tan(p)}}}{{{""=""}}}{{{"opposite"/"adjacent"}}}{{{""=""}}}{{{a/b}}}

The only side that is unknown is the adjacent side to angle "p", which is "b".

So we use the Pythagorean theorem, which is

{{{a^2+b^2}}}{{{""=""}}}{{{c^2}}} 

and substitute a=3 and c=5

{{{3^2+b^2}}}{{{""=""}}}{{{5^2}}}
{{{9+b^2}}}{{{""=""}}}{{{25}}}
{{{b²}}}{{{""=""}}}{{{25-9}}}
{{{b²}}}{{{""=""}}}{{{16}}}
{{{b}}}{{{""=""}}}{{{sqrt(16)}}}
{{{b}}}{{{""=""}}}{{{4}}}

Now we have side adjacent to angle "p", b=4
 
{{{drawing(200,500/3,0,6,0,5,
rectangle(4.7,1,5,1.3),
line(1,1,5,1), line(5,1,5,4), line(1,1,5,4),

locate(5.1,2.5,a=3),locate(2,2.9,c=5),locate(1.6,1.53,p), locate(3,1,"b=4")

 ) }}}

Now we can find tan(p):

{{{tan(p)}}}{{{""=""}}}{{{"opposite"/"adjacent"}}}{{{""=""}}}{{{a/b}}}{{{""=""}}}{{{3/4}}}

{{{tan(p)}}}{{{""=""}}}{{{3/4}}}

Edwin</pre>