Question

Open in App

Solution

Verified by Toppr

Correct option is B)

We know that:

$μ=sinA/2sin(2δm+A ) $

$⇒cotA/2=sin(A/2)sin(2δm+A ) $

$⇒sinA/2cosA/2 =sinA/2sin(2δm+A ) $

$⇒sin(π/2−A/2)=sin(2δm+A )$

$⇒π/2−A/2=2δm+A $

$⇒π−2A=δm$

Video Explanation

Was this answer helpful?

0

0