Question

to find the distance ab across a river, a surveyor laid off a distance bc = 351 m on one side of the river. it is found that b = 111°30 and c = 13°20. 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 \( A \)

The sum of angles in a triangle is \( 180^{\circ} \).
\( A=180^{\circ}-(111^{\circ}30'+13^{\circ}20') \)
\( 111^{\circ}30'+13^{\circ}20' = 124^{\circ}50' \)
\( A = 180^{\circ}-124^{\circ}50'=55^{\circ}10'\)
Convert \( 55^{\circ}10' \) to decimal: \( 55+\frac{10}{60}\approx55.17^{\circ}\), \( 111^{\circ}30'=111.5^{\circ}\), \( 13^{\circ}20'=13+\frac{20}{60}\approx13.33^{\circ}\)

Step2: Apply the Law of Sines

The Law of Sines states that \(\frac{AB}{\sin C}=\frac{BC}{\sin A}\)
We know \( BC = 351\) m, \( C = 13.33^{\circ}\), \( A=55.17^{\circ}\)
\( AB=\frac{BC\times\sin C}{\sin A}\)
Substitute the values: \( AB=\frac{351\times\sin(13.33^{\circ})}{\sin(55.17^{\circ})}\)
\(\sin(13.33^{\circ})\approx0.23\), \(\sin(55.17^{\circ})\approx0.82\)
\( AB=\frac{351\times0.23}{0.82}\)
\(351\times0.23 = 80.73\)
\( AB=\frac{80.73}{0.82}\approx98.45\approx98\)

Answer:

\(98\)