Question

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

Answer 1

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`

Answer 2

`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`