In the figure shown `sin i/sin r` is equal to

The power dissipated in the resistance R, in the circuit as shown in the figure, is maximum if R is equal to

A ray of light travels from a medium of refractive index n_(1) to a medium of refractive index n_(2) . If angle of incidence is i and angle of refraction is r, than (sin i)/(sin r) is equal to

Consider the circuit shown in the figure. The current i_(3) is equal to

Consider the circuit shown in the figure. The current I_(3) is equal to

Consider the circuit shown in the figure. The current I_(3) is equal to

When light is incident on a medium at angle i and refracted into a second medium at an angle r, the graph of sin i vs sin r is as shown in the graph, From this, one can conclude that

If sin 2A = 2 sin A, then A is equal to

In Delta ABC,2R^(2)sin A sin B sin C is equal to