Question

Three cities, A, B, and C, are located so that city A is due east of city B. If city C is located 35° west of north from city B and is 100 miles from city A and 70 milesfrom city B, how far is city A from city B?City Ais 20 miles due east of city B.City A is 35 miles due east of city B.City A is 42 miles due east of city B.City A is 122 miles due east of city B.

Answer 1

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.

`\begin{gathered} (AC)/(\sin B)=(BC)/(\sin A) \\ (100)/(\sin125\degree)=(70)/(\sin A) \\ \sin A=70\cdot(\sin 125\degree)/(100) \\ \sin A=0.573 \\ A=\sin ^(-1)(0.573) \\ A\approx35\degree \end{gathered}`

By the angle sum property of triangle the value of angle C can be calculated as,

`\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}`

Now, the distance between A and B can be calculated by,

`\begin{gathered} (AB)/(\sin C)=(BC)/(\sin A) \\ (AB)/(\sin20\degree)=(70)/(\sin 35\degree) \\ AB=\sin 20\degree\cdot(70)/(\sin 35\degree) \\ AB\approx42\text{ miles} \end{gathered}`

Thus, the distance of city A is 42 miles due east of city B.

Hence, option (C) is the correct answer.