Answer

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**Hint:**We are given with an equation and are asked to find the initial phase angle for the same. Thus, we will firstly evaluate the equation at time $ t = 0 $ . Then, we will use some basic trigonometric ideas to manipulate the evaluated value and then come up with an answer.

**Complete Step By Step Solution**

Here, The given equation is,

$ i = 10\sin \omega t + 8\cos \omega t $

Now, For the initial value, we take time $ t = 0 $

Taking here, we get

$ i = 10\sin \left( 0 \right) + 8\cos \left( 0 \right) $

We know,

$ \sin \left( 0 \right) = 0 $ And $ \cos \left( 0 \right) = 1 $

Thus, we get

$ i = 8\left( 1 \right) $

Further, we get

$ i = 8 $

Now,

$ {i_o} = \sqrt {{{\left( {10} \right)}^2} + {{\left( 8 \right)}^2}} $

Further, we get

$ {i_o} = \sqrt {164} $

Where, $ {i_o} $ is the amplitude of the motion.

Now,

As per the generic equation of such motion,

$ i = {i_o}\sin \left( {\omega t + \phi } \right) $

For time $ t = 0 $ ,

$ i = {i_0}\sin \phi $

Then, we get

$ \sin \phi = \dfrac{i}{{{i_o}}} $

Thus, we get

$ \sin \phi = \dfrac{8}{{\sqrt {164} }} $

Thus,

$ \tan \phi = \dfrac{8}{{\sqrt {164 - 64} }} $

Thus,

$ \tan \phi = \dfrac{8}{{10}} $

Thus,

$ \tan \phi = \dfrac{4}{5} $

Hence, we get

$ \phi = {\tan ^{ - 1}}\left( {\dfrac{4}{5}} \right) $

**Hence, the correct option is (A).**

**Note**

We have converted the sine function to a tangent one as all the given options are in the same format. We used basic trigonometry for conversion. One should not confuse it to be a given parameter.

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