Hyperboloid of one sheet as a ruled surface

Find points $A,B$ from system

$f(x,y,z)=\frac{x^2}{4}+ \frac{y^2}{25}- \frac{z^2}{9}-1=0$

$z=0$

$f_x(P_0)(x-P_{0x})+f_y(P_0)(y-P_{0y})+f_z(P_0)(z-P_{0z})=0$ - tangent plane in $P_0$

$P_0=(2 \sqrt{2}, 5 \sqrt{3}, 6)$

Here

$x_A = \frac{2}{5} (\sqrt{2} - 2 \sqrt{3}), y_A = 2 \sqrt{2} + \sqrt{3}, z_A = 0$

$x_B = \frac{2}{5} (\sqrt{2} + 2 \sqrt{3}), y_B = -2 \sqrt{2} + \sqrt{3}, z_B = 0$

And next find angles

$\angle P_0AP_p$ and $\angle P_0BP_p$ , $P_p=(2 \sqrt{2}, 5 \sqrt{3}, 0)$ .

First way

$\ cos \angle P_0AP_p = \frac {AP_0 \cdot AP_p }{\mid AP_0 \mid \cdot \mid AP_p \mid }$

$\ cos \angle P_0BP_p = \frac {BP_0 \cdot BP_p }{\mid BP_0 \mid \cdot \mid BP_p \mid }$

or second way from triangles $P_0AP_p, P_0BP_p$

$AP_0 \cdot \sin \angle P_0AP_p=P_0 P_p=6$

$BP_0 \cdot \sin \angle P_0BP_p=P_0 P_p=6$

Find angles

Wolfram 1

$\angle P_0AP_p=31.49 \degree$

Wolfram 2

$\angle P_0BP_p=47.55 \degree$

Answer $79.04 \degree$ .

P.S. GEOGEBRA 3D