asked 160k views
5 votes
Find cosα given that sinα = 5/13 and α is in quadrant II

asked
User JoeBobs
by
7.7k points

1 Answer

5 votes

Answer:

cos α = -
(12)/(13)

Explanation:

Since α is in second quadrant then cosα < 0

Using the identity

sin²α + cos²α = 1 then

cosα = -
√(1-sin^2\alpha )

= -
\sqrt{1-((5)/(13))^2 }

= -
\sqrt{1-(25)/(169) }

= -
\sqrt{(144)/(169) }

= -
(12)/(13)

answered
User Rickerby
by
8.4k points
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