Question

to find the distance ab across a river, a surveyor laid off a distance bc = 352 m on one side of the river. it is found that b = 113°30 and c = 14°10. find ab. the distance ab across the river is \boxed{} m (simplify your answer. do not round until the final answer. then round to the nearest whole number as needed.)

Explanation:

Step1: Find angle \(A\)

In \(\triangle ABC\), the sum of angles is \(180^{\circ}\).
\(A = 180^{\circ}-(113^{\circ}30'+ 14^{\circ}10')\)
\(A = 180^{\circ}-127^{\circ}40'\)
\(A = 52^{\circ}20'\)
Convert \(A = 52^{\circ}20'=\frac{52\times60 + 20}{60}=\frac{3120+20}{60}=\frac{3140}{60}\approx52.33^{\circ}\), \(B = 113^{\circ}30'=\frac{113\times60+30}{60}=\frac{6780 + 30}{60}=\frac{6810}{60}=113.5^{\circ}\), \(C = 14^{\circ}10'=\frac{14\times60+10}{60}=\frac{840+10}{60}=\frac{850}{60}\approx14.17^{\circ}\)

Step2: Apply the Law of Sines

By the Law of Sines \(\frac{AB}{\sin C}=\frac{BC}{\sin A}\)
We know \(BC = 352\) m, \(\sin C=\sin(14^{\circ}10')\approx\sin(14.17^{\circ})\approx0.245\), \(\sin A=\sin(52^{\circ}20')\approx\sin(52.33^{\circ})\approx0.791\)
Substitute into the formula: \(AB=\frac{BC\times\sin C}{\sin A}\)
\(AB=\frac{352\times\sin(14^{\circ}10')}{\sin(52^{\circ}20')}\)
\(AB=\frac{352\times0.245}{0.791}\)
\(AB=\frac{86.24}{0.791}\approx109\)

Answer:

\(109\)