This is hardly an astronomy question, but I like drawing, so here you are:
Longitudes are measured from the Greenwich meridian, so the angle between Kampala and Quito is
$$theta_mathrm{Q} + theta_mathrm{K} = 82.5^circ + 37.5^circ = 110^circ.$$
(remember that $0^circ30' = 0.5^circ$).
The shortest surface path is along Equator. Since $110^circ$ is $frac{110^circ}{360^circ} simeq 0.3$ times the circumference of Earth at Equator, the length of path A (the dashed line) is
$$mathrm{A}:,,d = 0.3 times 40,075,mathrm{km} = 12,245,mathrm{km}$$
For path B (the solid line), you need a bit more trigonometry. The radius of Earth is $R = 40,075,mathrm{km},/,2pi = 6,378,mathrm{km}$. The rest will be left as an exercise.
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