Question

A ship sails on course N30oE finds a lighthouse bearing 120o true, 5
miles away.


1.) How far should the ship sails to find the lighthouse due south?

Answer 1

A= $x$

C = 30$\degree$

B= 120$\degree$

A = 180$\degree$ - 30$\degree$ - 120$\degree$

A = 30$\degree$

$\frac{sinA}{5miles}$ = $\frac{sinB}{b}$

$\frac{sin30}{5miles}$ = $\frac{sin120}{b}$

$\frac{1/2}{5miles}$ = $\frac{\sqrt3/2}{b}$

1/2 ( b )= $\sqrt3/2$ (5miles)

b = 5 $\sqrt3$ miles

the ship sails to find the lighthouse due south is 5$\sqrt3$ miles