Secants of a circle

If two secants intersect outside the circle, the product of the measures of one secant and its external segment equals the product of the measures of the other secant and its external segment. Thus, $(PA)(PB)=(PD)(PC)$ . Substituting, we get

$(PA)(33)=(3)(77)$ $\implies$ $PA=7$

Therefore, $BA=33-7=26$ .

A line perpendicular to a chord of a circle and containing the center of the circle, bisects the chord and its major and minor arcs. Thus,

$BE=AE=13$

In right $\triangle PEO$ , $PO=40$ and $PE=20$ .

It follows that,

$\cos \angle BPC=\dfrac{20}{40}$ $\color{#D61F06}\large \implies$ $\angle BPC =\cos^{-1}(\frac{1}{2})=60^{\circ}$

Triangle $OPB$ has sides

$BP=33$

$OB=\dfrac12\times74=37$

$OP=37+3=40$

From a law of cosines

$\cos(\angle OPB)=\dfrac{40^2+33^2-37^2}{2\times40\times33}=\dfrac12$

$\angle OPB=\arccos\dfrac12=\boxed{60^\circ}$

$(DP)=(CP)-(CD)=77-74=3$
$(OD)=\frac{(CD)}{2}=37$
$(OA)=(OD)=37$
$(OP)=(OD)+(DP)=37+3=40$
$(PA)(PB)=(PD)(PC)\Rightarrow (PA)\cdot 33=3\cdot 77\Rightarrow (PA)=7$
If $\theta$ is the angle we are looking for then by applying the cosine law on the triangle OAP we get:
$(OA)^{2}=(OP)^{2}+(PA)^{2}-2\cdot (OP)(PA)\cos{\theta}\Rightarrow$
$1369=1600+9-2\cdot 40\cdot 7\cdot \cos{\theta}\Rightarrow$
$\cos{\theta}=\frac{1}{2}\Rightarrow$
$\theta=60^{\circ}$