What's the measure of the $\angle POB$ in the circle below?

For reference:

If you have a circle with center $O$ and radius $\sqrt5 + 1$. through an outer point $P$, the secant $PBA $and the tangent $PE$ are traced. if $PE = AB = 2$, calculate $\angle POB$.(Answer:$18^\circ$)

My progress:

$\triangle ABO(\text{isosceles}) : AB=2, OA=OB = \sqrt5+1 \implies\\ \angle POA = 36^o, \angle OAB=\angle ABO = 72^o\\ \triangle POE: 2^2+(\sqrt5+1)^2 = OP^2 \implies OP = \sqrt{10+2\sqrt5}$

Th.Stweart: $\triangle OAP: \\ OA^2.BP+OP^2.2 = BO^2.(2+BP)+(2+BP).2.BP\\ \therefore BP = \sqrt5-1$

Law of cosines $\triangle POB: \\ BP^2=BO^2+OP^2-2.BO.OP.cos \angle BOP \implies\\ \boxed{\angle BOP=18^o}$

Is the solution possible without using the law of cosines since it results in an unusual angle?

Using power of the point $P$ with respect to circle $O$,

$PB \cdot PA = PE^2 \implies PB \cdot (PB +2) = 4$

That gives $PB = \sqrt5 - 1$ and hence $PA = \sqrt5 + 1$

As $PA = OA = \sqrt5 + 1, ~ \angle AOP = \angle APO = 54^\circ$

$ \therefore \angle BOP = 18^\circ$