Solution:
A girl start at A and walk 2km south to B she then walk 3km west to C.
The diagrammatic representation is shown below
To find the distance of C from A "AC", we apply the Pythagorean theorem formula which is
[tex]\begin{gathered} (Hypotenuse)^2=(Opposite)^2+(Adjacent)^2 \\ Opposite=2\text{ km} \\ Adjacent=3\text{ km} \\ Hypotenuse=AC \end{gathered}[/tex]Substitute the values into the formula above
[tex]\begin{gathered} (AC)^2=2^2+3^2=4+9=13 \\ (AC)^2=13 \\ Square\text{ root of both sides} \\ \sqrt{(AC)^2}=\sqrt{13}\text{ km} \\ AC=3.61\text{ km \lparen nearest hundredth\rparen} \end{gathered}[/tex]Hence, the distance of C from A is 3.61 km (nearest hundredth)
To find the bearing of C from A, we find the value of θ,
Applying SOHCAHTOA
[tex]\begin{gathered} \tan\theta=\frac{Opposite}{Adjacent}=\frac{2}{3} \\ \theta=\tan^{-1}(0.6667) \\ \emptyset=33.69\degree\text{ \lparen nearest hundredth\rparen} \end{gathered}[/tex]The bearing of C from A will be
[tex]\begin{gathered} Bearing\text{ of C from A}=180\degree+(90\degree-\theta) \\ Bearing\text{ of C from A}=180\degree+90\degree-33.69\degree=236.31\degree \\ Bearing\text{ of C from A}=236.31\degree \end{gathered}[/tex]Hence, the bearing of C from A is 236.31°
