Question

Question #ba73b

Answer 1

In the above right angled triangle `/_BCA+/_CBA=90^@`

So`" "/_BCA+/_CBA=90^@`

`=/_CBA=90^@-/_BCA`

Now
`sin/_BCA=c/a=cos/_CBA`

`=>sin/_BCA=cos(90^@-/_BCA)`

`=>sin30^@=cos(90^@-30^@)`