Question

Question #7ebbf

Answer 1

The value is `8.83`

Explanation:

Given `sin20^@.cos70^@`
For solving by multiplying and dividing by 2 we have
`1/2.(2sin20^@.cos70^@)`
It is of the form `2sinAcosB` but `2sinAcosB=sin(A+B)+sin(A-B)`
Substituting we get `:`
`1/2(2sin90^@.sin50^@)rArr1/2(1+0.766)rArr1/2(1.766)rArr8.83`