Melody

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   :)))

$$\\44-2(3x+4)=-18x\\\\
44-6x-8=-18x\\\\
36-6x=-18x\\\\
36-6x+6x=-18x+6x\\\\
36=-12x\\\\
-3=x\\\\
x=-3$$

$$6x(3x-4)=18x^2-24x$$

inverse tan is the same as arc tan so use atan     (also use acos and asin)

Well

$$\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right) = {\frac{{\mathtt{376}}}{{\mathtt{15}}}} = {\mathtt{25.066\: \!666\: \!666\: \!666\: \!666\: \!7}}$$

$$\frac{37625}{15} = 25\frac{1}{25}$$

so   the quotient of         $$25\frac{1}{15}\quad and \quad 7\frac{5}{6}\quad is \quad 3\frac{1}{5}$$

check:

$${\frac{\left({\mathtt{25}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{15}}}}\right)}{\left({\mathtt{7}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{5}}}{{\mathtt{6}}}}\right)}} = {\frac{{\mathtt{16}}}{{\mathtt{5}}}} = {\mathtt{3.2}}$$

3 1/4 divided by 1 1/3

(3+ 1/4 ) / (1+ 1/3)

$${\frac{\left({\mathtt{3}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}} = {\frac{{\mathtt{39}}}{{\mathtt{16}}}} = {\mathtt{2.437\: \!5}}$$

(16y−4z8)^1/4 /(32y5z−10)^−1/5

$$(16y-4z8)^{1/4} *(32y5z-10)^{1/5}$$

Here is the first on but you need to label the axes

See if this helps :)

That is a really interesting question.   And a great answer  

I wish I could give you more points Heureka