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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,

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

Now,

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

Taking here, we get

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

We know,

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

Thus, we get

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

Further, we get

$i = 8$

Now,

$\Rightarrow$ ${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,

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

Thus,

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

Thus,

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

Hence, we get

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

**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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