Trouble with angles and lengths

Our first target is to find out another side of $\triangle PQR$
In $\triangle ABP$ ,
$\angle APB = 180^{\circ} -(30^{\circ} + 60^{\circ})$ = $90^{\circ}$
Hence, $\cos 30^{\circ} = \frac {BP}{200}$
$BP = 200 \cos 30^{\circ} = 173.2m$

Again, In $\triangle ABQ$ ,
Hence, $\frac {BQ}{\sin 46^{\circ}}= \frac {200}{\sin 67^{\circ}}$
$BQ = 156.3m$

Then in $\triangle PBQ$ ,the cosine rule gives,
$PQ^{2} = BP^{2} + BQ^{2} - 2*BP*BQ \cos97^ {\circ }$ $= (173.2)^{2} + (156.3)^{2} - 2(173.2)(156.3)(-.1219)$
Hence, $PQ = 247m$

Now,In $\triangle PQR$ ,
$\frac{266.36}{\sin 80.6^{\circ}} = \frac{247}{\sin PRQ}$
or, $\sin PQR = .91486455$
or, $\angle PQR = \sin^{-1} ({ .91486455 })$
so, $\angle PQR = 66.19^{\circ}$
Again $\triangle PQR$ ,
$\angle RPQ = 180^{\circ} - (80.6^{\circ} + 66.19^{\circ})$
so, $\angle RPQ = 33.21^{\circ}$

Now,In $\triangle PQR$ ,
$\frac{266.36}{\sin 80.6^{\circ}} = \frac{RQ}{\sin 33.21^{\circ} }$
so, $RQ = 147.87$
The area of the $\triangle PQR$ = $1/2 * r*p* \sin PQR$
= $1/2 * 247*147.87* \sin 80.6^{\circ}$
= $18016.73$
Let $x$ is the distance of $O$ from $p$ or $q$ or $r$
Area of $\triangle POQ = 1/2* r*x$
Area of $\triangle ROP = 1/2* q*x$
Area of $\triangle ROQ = 1/2* p*x$

Thus the total area of $\triangle PQR = 1/2*x(p+q+r)=300.615*x$
But this area is $18016.73$ . Herce, the area of $O$ from $RQ , x= \frac {18016.73}{300.615} =\boxed{54.5}$
We can write $a<54.5<a+1$ $54<54.5<54+1$ or $54<54.5<55$
SO, $a= \boxed{54}$