Question

if sin tita = 4/5 and tita is not in first quardent, find the valu of cos tita?

Answer 1

tita belongs to second quadrant

Answer 2

sin thita =perpendicular/hypotenuse
= 4x/5x
=4/5
p=4;h=5
let AB=perpendicular ; BC=base ; AC=hypotenuse
by Pythagoras theorem

$bc= \sqrt{ac {}^{2} - ab {}^{2} }$
$\: \: = \sqrt{5 {}^{2} - 4 {}^{2} }$
$\: \: = \ \sqrt{25 - 16}$
$= \sqrt{9}$
$= 3$
we get ac=hypotenuse=5
bc=base=3
ab=perpendicular=4
cos thita = base/hypotenuse
=3/5
thita belongs to the second quadrant