# Justice starts a 5km walk from P on a bearing 023. He then walks 4km on a bearing of 113 to Q. What is the bearing of Q from P?.

May 26, 2018

${61}^{\circ} 39 ' 35 ' '$ $\text{Bearing P-Q}$

#### Explanation:

$\text{Let the turning point be R}$

$\therefore R - Q = {113}^{\circ}$ $\text{Given}$

$\therefore P - R = {23}^{\circ}$ $\text{Given}$

$\therefore R - P = {23}^{\circ} + {180}^{\circ} = {203}^{\circ}$

$\therefore {203}^{\circ} - {113}^{\circ} = {90}^{\circ} = \angle R$

$\therefore \text{in right}$ $\triangle$ $\text{QPR}$$: -$

$\therefore \tan \angle P = \frac{4}{5} = 0.8 \arctan = {38}^{\circ} 39 ' 35 ' ' = \angle P$

$\therefore {180}^{\circ} - \left({90}^{\circ} + {38}^{\circ} 39 ' 35 ' '\right) = {51}^{\circ} 20 ' 25 ' ' = \angle Q$

$\therefore {113}^{\circ} + {180}^{\circ} = {293}^{\circ} = \text{ bearing Q-R}$

$\therefore {293}^{\circ} - {51}^{\circ} 20 ' 25 ' ' = {241}^{\circ} 39 ' 35 ' ' = \text{bearing Q-P}$

$\therefore \text{bearing P-Q"=241^@39'35''-180^@=61^@39'35''="bearing P-Q}$