lenght of the shadow of a 15m high pole is 5sqrt3 m at 7o’ clock in the morning . then what is the angle of elevation of sun rays with the ground at the time

Let us take the scenario in the form of a right triangle, where the height of the pole is represented as AB, the shadow of the pole is represented as BC. Let 
∠ C be the angle of Elevation.

So, We know that,

Tan Ф = Opposite / Adjacent

=> Tan Ф = 15 / 15 √ 3

=> Tan Ф = 1 / √ 3

We know that, Tan 30° = 1 / √ 3

So we can say that, Ф = 30°

Hence the angle of Elevation of the sum at that time was 30°.