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

2 Answers
Jun 4, 2018

Answer:

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

Answer:

#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#