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.
Use
sin²A+cos²A=1
so
cosA = sqrt(1-sin²A)
=sqrt(1-(10/13)²)
=sqrt(69/169)
=sqrt(69)/13
What is the value of cos A???
please because i forget.
sin²A+cos²A=1
so
cosA = sqrt(1-sin²A)
=sqrt(1-(10/13)²)
=sqrt(69/169)
=sqrt(69)/13