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$\angle A = x,\angle B = 3x,\angle C = 5x$ and use the sum of angles $ = {180^ \circ }$, you will get the answer.

So, basically, the question is saying that there is a triangle whose angles are in ratio $1:3:5$. And now we have to find each angle.

So first of all, let us assume that $ABC$ is a triangle.

Now, it is given that angles are in ratio $1:3:5$

So, we can write $\angle A,\angle B,\angle C$ as the angles of $\vartriangle ABC$.

Now according to question, it is given that

$\angle A:\angle B:\angle C = 1:3:5$

As we know that if we remove the proportionality sign we must multiply by a constant.

Let $x$ be the constant

Then we can say that

$

\angle A = 1 \times x \\

\angle B = 3 \times x \\

\angle C = 5 \times x \\

$

So, we get all three angle in terms of $x$

That is, $\angle A = x,\angle B = 3x,\angle C = 5x$

So as we all know that the sum of angles of a triangle is ${180^ \circ }$.

i.e. $\angle A + \angle B + \angle C = {180^ \circ }$

Here we put the value of all three angles , we will get

$

x + 3x + 5x = 180 \\

9x = 180 \\

x = {20^ \circ } \\

$

So now we can easily find the values of all the angles.

$

\angle A = x = {20^ \circ } \\

\angle B = 3x = 3 \times 20 = {60^ \circ } \\

\angle C = 5x = 5 \times 20 = {100^ \circ } \\

$

(As we know that $\angle A = x,\angle B = 3x,\angle C = 5x$)

Hence, ${20^ \circ },{60^ \circ },{100^ \circ }$ are the angles of the triangle.

This means, $\dfrac{{\angle A}}{{\angle B}} = \dfrac{1}{3},\dfrac{{\angle B}}{{\angle C}} = \dfrac{3}{5}$

So, $

\angle A = x \\

\angle B = 3x \\

\angle C = \dfrac{5}{3}\angle B \\

$

That is, $\angle C = \dfrac{5}{3} \times 3x = 5x$

Then we put in the equation of $\vartriangle ABC$

$\angle A + \angle B + \angle C = {180^ \circ }$

And we will get the required answer.