Question

Four ships A, B, C and D are at sea in the following relative positions : B is on the straight line segment AC, B is due north of D, and D is due west of C. The distance between B and D is 2 km, $\angle BDA={40}^o,\angle BCD={25}^o$. What is the distance between A and D? (Take $sin{25}^o=0.423$).

Answer 1

The figure is self explanatory. In $\Delta ADC$, we have $\angle ADC={40}^o+{90}^o={130}^o$
and $\angle DCA={25}^o$
$\therefore \angle DAC={180}^o-({130}^o+{25}^o)={25}^o$
Hence $AD=DC=2\cot {25}^o=2.\frac{\sqrt{1-{\sin }^2{25}^o}}{\sin {25}^o}$
$=\frac{2\times .906}{0.423}=4.283km$.