Given:
City A is due east of city B.
City C is located 35° west of north from city B.
Distance between city C and city A is 100 miles.
Distance between city C and city B is 70 miles.
The objective is to find the distance between city A and city B.
The above situation can be represented as,
Thus the total angle of ∠B = 90°+35° = 125°.
Now the measure of angle A can be calculated by law of sines.
[tex]\begin{gathered} \frac{AC}{\sin B}=\frac{BC}{\sin A} \\ \frac{100}{\sin125\degree}=\frac{70}{\sin A} \\ \sin A=70\cdot\frac{\sin 125\degree}{100} \\ \sin A=0.573 \\ A=\sin ^{-1}(0.573) \\ A\approx35\degree \end{gathered}[/tex]By the angle sum property of triangle the value of angle C can be calculated as,
[tex]\begin{gathered} \angle A+\angle B+\angle C=180\degree \\ 35\degree+125\degree+\angle C=180\degree \\ \angle C=180\degree-35\degree-125\degree \\ \angle C=20\degree \end{gathered}[/tex]Now, the distance between A and B can be calculated by,
[tex]\begin{gathered} \frac{AB}{\sin C}=\frac{BC}{\sin A} \\ \frac{AB}{\sin20\degree}=\frac{70}{\sin 35\degree} \\ AB=\sin 20\degree\cdot\frac{70}{\sin 35\degree} \\ AB\approx42\text{ miles} \end{gathered}[/tex]Thus, the distance of city A is 42 miles due east of city B.
Hence, option (C) is the correct answer.
