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For the given triangle ABC, the angle $\angle A=\theta $ and angle $\angle C={{90}^{\circ }}$. We can infer that the triangle is a right-angled triangle.

For the given angle $\theta $, we can write that AC = a is the adjacent side, BC = b is the opposite side and AB = c is the hypotenuse.

From the basic definitions of trigonometric t=ratios, we can write that

$\tan \theta =\dfrac{\text{Opposite side}}{\text{Adjacent side}}=\dfrac{b}{a}$

$\sec \theta =\dfrac{\text{Hypotenuse}}{\text{Adjacent side}}=\dfrac{c}{a}$

Let us consider the triangle OAB from the above diagram,

As given in the question, we can write that angle $\angle A={{90}^{\circ }}$, the triangle is a right angled triangle. For the given angle $\theta $, we can write that the side AB is opposite side, the side OA is adjacent side and OB is hypotenuse. We can write that

$\begin{align}

& \tan \theta =\dfrac{\text{Opposite side}}{\text{Adjacent side}}=\dfrac{AB}{OA}=\dfrac{AB}{r} \\

& AB=r\tan \theta \to \left( 1 \right) \\

\end{align}$

$\begin{align}

& \sec \theta =\dfrac{\text{Hypotenuse}}{\text{Adjacent side}}=\dfrac{OB}{OA}=\dfrac{OB}{r} \\

& OB=r\sec \theta \to \left( 2 \right) \\

\end{align}$

The length of the arc of a circle of radius r and the included angle $\theta $ is given by the formula $\text{arc length}=\dfrac{\theta }{{{360}^{\circ }}}\times 2\pi r$.

The length of the arc PA which includes an angle $\theta $ at the centre is

$PA=\dfrac{\theta }{{{360}^{\circ }}}\times 2\pi r=\dfrac{{{\theta }^{\circ }}}{180}\times \pi r\to \left( 3 \right)$

We can write the length of the side PB as $PB=OB-OP$

As we know that OP = r, we can write that

$PB=OB-r\to \left( 4 \right)$

We know that the required perimeter is the sum of AB, PB, arc PA. Mathematically,

$Perimeter=AB+PB+arc\left( PA \right)$

From the equations 1, 2, 3, 4, we can write that

$\begin{align}

& Perimeter=r\tan \theta +OB-r+\dfrac{{{\theta }^{\circ }}}{180}\times \pi r \\

& Perimeter=r\tan \theta +r\sec \theta -r+\dfrac{{{\theta }^{\circ }}}{180}\times \pi r \\

\end{align}$

By taking r common in the whole R.H.S, we get that

$\begin{align}

& Perimeter=r\left( \tan \theta +\sec \theta +\dfrac{{{\theta }^{\circ }}}{180}\times \pi -1 \right) \\

& Perimeter=r\left( \tan \theta +\sec \theta +\dfrac{\pi {{\theta }^{\circ }}}{180}-1 \right) \\

\end{align}$