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## Main Question or Discussion Point

Say I have an acute angle A in standard position in the first quadrant with a rotating arm of length r , terminating at coordinate P(a,b) . Now say I rotate it 90 degrees further from that position to the second quadrant , this ends up at coordinate Q(-a,b) . I draw perpendiculars from those two positions to the x axis .

for the first triangle , sin A = b/r and cos A = a/r

as for the second triangle , why is sin (90+A) = a/r = cos A

and why is cos(90+A) = -b/r = -sin A

Im confused ... how can you compute the trig ratio for an angle greater than 90 degress ? Why not take the ratio for the acute angle (180 - (90+a))

Pls Help .

for the first triangle , sin A = b/r and cos A = a/r

as for the second triangle , why is sin (90+A) = a/r = cos A

and why is cos(90+A) = -b/r = -sin A

Im confused ... how can you compute the trig ratio for an angle greater than 90 degress ? Why not take the ratio for the acute angle (180 - (90+a))

Pls Help .