If sin A = 1/3 and 0< A < 90, then what is cos A?

2 Answers
Jun 4, 2018

See below

Explanation:

If sin A=1/3 and 0 < A < 90

Then using sin^2 A+cos^2A=1

cos^2A=1-sin^2A=1-1/9=8/9

Then cosA=+-sqrt(8/9)=+-(2sqrt2)/3

But if 0< A< 90, then the cosine is positive and cosA=(2sqrt2)/3

Jun 4, 2018

cos A = (2sqrt2)/3

Explanation:

Since 0 < A < 90^o we know that angle A is in the first quadrant.

Consider the right triangle OAB where O is the origin and side AB is opposite angle A

We are told that sin A =1/3

:. side AB =1 and side OA =3

Applying Pythagoras

(OA)^2 = (AB)^2 + (OB)^2

:. 3^2 = 1^2 +OB^2

OB = sqrt(9-1) [Since OB>0 because angle A is in the first quadrant]

OB = sqrt8 = 2sqrt2

Now, cos A = (OB)/(OA)

cos A = (2sqrt2)/3