Let the point (x,y ) be the point where the speed boat intercepts the vessel
Let t be the time taken to intercept.
Let alpha be the angle between East and the direction the speed boat travels (as shown in the diagram)
So the bearing of the speed boat will be N(90-alpha)E

$$\\x=20cos75+26tcos50\qquad and \qquad x=50tcos\alpha\\
so\\
20cos75+26tcos50=50tcos\alpha\\\\
20cos75=50tcos\alpha-26tcos50\\\\
20cos75=t(50cos\alpha-26cos50)\\\\
\frac{20cos75}{ (50cos\alpha-26cos50)}=t\\\\
AND\\
y=20sin75+26tsin50\qquad and \qquad y=50tsin\alpha\\\\
...\\
\frac{20sin75}{ (50sin\alpha-26sin50)}=t\\
so\\
\frac{(20sin75)}{ (50sin\alpha-26sin50)}=\frac{(20cos75)}{ (50cos\alpha-26cos50)}\\\\$$
$$\\(20sin75)(50cos\alpha-26cos50)=(20cos75)(50sin\alpha-26sin50)\\\\
(20sin75)(50cos\alpha)-(20sin75)(26cos50)=(20cos75)(50sin\alpha)-(20cos75)(26sin50)\\\\
(1000sin75)(cos\alpha)-(520sin75cos50)=(1000cos75sin\alpha)-(520cos75sin50)$$
$$\\(1000sin75cos\alpha)-(1000cos75sin\alpha)=(520sin75cos50)-(520cos75sin50)\\\\
1000(sin75cos\alpha-cos75sin\alpha)=520(sin75cos50-cos75sin50)\\\\
1000sin(75-\alpha)=520sin(75-50)\\\\
sin(75-\alpha)=0.52sin(25)\\\\
75-\alpha=asin(0.52sin(25))\\\\
-\alpha=asin(0.52sin(25))-75\\\\
\alpha=75-asin(0.52sin(25))\\\\$$
$${\mathtt{75}}{\mathtt{\,-\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\mathtt{0.52}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{25}}^\circ\right)}\right)} = {\mathtt{62.304\: \!975\: \!095\: \!575}}$$
$${\mathtt{90}}{\mathtt{\,-\,}}{\mathtt{62.304\: \!975\: \!095\: \!575}} = {\mathtt{27.695\: \!024\: \!904\: \!425}}$$ = 27 degrees 41minutes and 42 seconds
The speed boat needs to set a course for
$$N\;27^041'42"E$$
What a great question and Heureka and I even agree :)))