Answer:
24.3°
Step-by-step explanation:
A suitable calculator can add the vectors directly. (See attached.) In this calculation, bearing angles are measured clockwise from north, so the initial vector has an associated angle of 0 degrees.
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We can find the angle of interest by considering the distances north and east that city C is from city B. The relevant trig functions are ...
Sin = Opposite/Hypotenuse
Cos = Adjacent/Hypotenuse
Tan = Opposite/Adjacent
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location of C relative to B
The direct distance from B to C is the hypotenuse of a right triangle whose right angle is at point "D" due north of B and due west of C. The distances of interest are ...
sin(57°) = DC/BC
DC = BC·sin(57°) = (108 km)sin(57°) ≈ 90.5764 km
cos(57°) = BD/BC
BD = BC·cos(57°) = (108 km)cos(57°) ≈ 58.8210 km
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angle of C from A
And the angle of interest can be found from ...
tan(θ) = DC/AD = (90.5764 km)/(142 km +58.8210 km) ≈ 0.451031
θ = arctan(0.451031) ≈ 24.3°
The pilot could have flown on a bearing of 24.3° to get to city C directly from city A.
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Additional comment
We can ignore the fact that bearing angles are measured CW from north, while conventional angles on a Cartesian plane are measured CCW from "east" (+x). Each coordinate system can be mapped to the other by flipping the map over along a NE-SW line (y=x), aligning the +x axis with north and the +y axis with east.
The diagram we used in working this problem is shown in the second attachment.
