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Asked by susi

In triangle ABC, if <A and <B are acute angles, and sin A 10/13 .
What is the value of cos A???

please because i forget.
14 years ago

Answers

Answered by MathMate
Use
sin²A+cos²A=1
so
cosA = sqrt(1-sin²A)
=sqrt(1-(10/13)²)
=sqrt(69/169)
=sqrt(69)/13
14 years ago

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