Question

sinA
=a/b
then coso is ?​

Answer 1

see attachement....

Khushi here✔️✌

Answer 2

Answer:

Here is your answer

Step-by-step explanation:

instead of theta I'm using x

sinx = a/b

cosx = √1-sin^2x

=√{1-(a/b)^2}

=√{(b^2-a^2)/b^2

=√{(b^2-a^2)/b^2

=√(b^2-a^2)/b

Hope this will help you