Chapter 8 - Application of Trigonometry - Summary Exercises on Applications of Trigonometry and Vectors - Exercises - Page 792: 3

$5856\ m$

1. Draw a diagram as shown. 2. We have angle $B=39^\circ43'=39.72^\circ$, angle $\angle CAD=129^\circ43'=129.72^\circ$. Thus angle $C=129.72^\circ-39.72^\circ=90^\circ$ 3. In the right triangle, $sinB=\frac{3742}{x}$, thus $x=\frac{3742}{sin39.72^\circ}\approx5856\ m$