Trigonometry problem solving in diagram A is a point on the $x$ axis and B is a point on the $y$ axis such that $(5,3)$ lies on the straight line passing through A and B . Given that OP is perpendicular to AB and $\angle BAO = \theta $ , Show that OP = $3 \ cos \theta + 5 \sin \theta $ 

My attempt , 
$\sin \theta = \frac{OP}{OA} $ 
$OP =  OA \sin \theta$
OA = $ 5 + ? $ 
The '?' is what I marked on the diagram as well. I'm not sure how do I get that .
 A: $$\tan \theta = \frac{3}{OA-5} $$
$$OA-5=3\cot \theta$$
$$OA=3\cot \theta+5$$
Thus
$$\sin \theta= \frac{OP}{OA} $$
$$OP=OA\sin \theta =3\cos \theta+5\sin \theta$$
A: 
\begin{align} 
|OA|&=5+3\cot\theta
,\\
|OB|&=|OA|\tan\theta
,\\
|OP|&=|OB|\cos\theta
=(5+3\cot\theta)\tan\theta \cos\theta
=(5+3\cot\theta)\sin\theta
=5\sin\theta+3\cos\theta
.
\end{align}  
A: The equation of $AB$
$$\tan(180^\circ-\theta)=\dfrac{y-3}{x-5}$$
$$\iff\dfrac x{\dfrac{3\cos\theta+4\sin\theta}{\sin\theta}}+\dfrac y{\dfrac{3\cos\theta+4\sin\theta}{\cos\theta}}=1$$
$OA=\dfrac{3\cos\theta+4\sin\theta}{\sin\theta}$
A: The normal form of the equation of a straight line is
        $$    x cos α + y sin α - p = 0     $$
where $\alpha$ is the angle from the x-axis to the perpendicular from the origin to the line and p is the length of the perpendicular from origin to the line.
So, here, $\alpha=90-\theta$
So, put in the equation and also the length of the perpendicular to the line from the origin=$OP$ and also $(5,3)$ lies on the line. So,
$$    5 \cos (90-\theta) + 3 \sin (90-\theta) - p = 0     $$ 
$$\implies 5\sin \theta +3 \cos \theta=OP$$
