Question 42331
{{{drawing(400,400,-10,10,-10,10,
line(0,0,6,-8),
green(line(6,0,6,-8)),
grid(1),
locate(6,-8,P),
locate(6,0,A),
locate(0,0,O)
)}}}


The coordinates of P are (6,-8).
AO = 6 units and AP = -8 units


From Pythagorus' theorem in rightangled triangle OAP,
{{{OP^2=AO^2 + AP^2}}}
or {{{OP=sqrt(AO^2 + AP^2)}}}
or {{{OP=sqrt(6^2 + (-8)^2)}}}
or {{{OP=sqrt(36 + 64)}}}
or {{{OP=sqrt(100)}}}
So, OP = 10 units


Now, cos &#952 = cos(< AOP) = {{{OA/OP}}} = {{{6/10}}} = {{{3/5}}}


Note: 
sin &#952 = sin(< AOP) = {{{AP/OP}}} = {{{(-8)/10}}} = {{{-4/5}}} and tan &#952 = tan(< AOP) = {{{AP/OA}}} = {{{(-8)/6}}} = {{{-4/3}}}