Question

A boy A is standing 20√3 m away in a direction 30° north of east from his friend B. Another boy C
standing somewhere east of B can reach A, if he walks in a direction 60° north of east. In a Cartesian
coordinate system with its x-axis towards the east, the position of C with respect to A is

​

Answer 1

Answer: 20 m west of south

Explanation:

We have,

Position of A at a distance of 20√3 m at 30° North of East of B.

Also, by 60° North of East of C.

So, we made a figure as mentioned in question.

In triangle ABC,

AB = 20√3 m

By using linear pair,

$\angle ABC = 30\degree \\ \\ \angle ACB +\angle ACO = 180\degree \\ \\ => \angle ACB + 60\degree =180 \degree \\ \\=>\angle ACB = 120\degree$

By applying sine rule,

$\frac{AC}{sin(\angle ABC )}=\frac{AB}{sin(\angle ACB)} \\ \\ \frac{AC}{sin(30°)}=\frac{AB}{sin(120\degree)} \\ \\ =>\frac{AC}{1/2}=\frac{20\sqrt{3}}{\sqrt{3}/2}\\ \\=>AC = 20m$

By angle sum property of triangle,

$\angle ACO + \angle OAC + \angle AOC = 180\degree \\ \\=>60\degree + \angle OAC + 90\degree = 180\degree \\ \\=>\angle OAC = 30\degree$

Hence, we get position of C with respect to A is 20m west of south at angle of 30° .

Answer 2

Answer:

Answer: 20 m west of south

Explanation:

We have,

Position of A at a distance of 20√3 m at 30° North of East of B.

Also, by 60° North of East of C.

So, we made a figure as mentioned in question.

In triangle ABC,

AB = 20√3 m

By using linear pair,

\begin{gathered}\angle ABC = 30\degree \\ \\ \angle ACB +\angle ACO = 180\degree \\ \\ = > \angle ACB + 60\degree =180 \degree \\ \\= > \angle ACB = 120\degree \end{gathered}∠ABC=30°∠ACB+∠ACO=180°=>∠ACB+60°=180°=>∠ACB=120°

By applying sine rule,

\begin{gathered} \frac{AC}{sin(\angle ABC )}=\frac{AB}{sin(\angle ACB)} \\ \\ \frac{AC}{sin(30°)}=\frac{AB}{sin(120\degree)} \\ \\ = > \frac{AC}{1/2}=\frac{20\sqrt{3}}{\sqrt{3}/2}\\ \\= > AC = 20m\end{gathered}sin(∠ABC)AC=sin(∠ACB)ABsin(30°)AC=sin(120°)AB=>1/2AC=3/2203=>AC=20m

By angle sum property of triangle,

\begin{gathered} \angle ACO + \angle OAC + \angle AOC = 180\degree \\ \\= > 60\degree + \angle OAC + 90\degree = 180\degree \\ \\= > \angle OAC = 30\degree \end{gathered}∠ACO+∠OAC+∠AOC=180°=>60°+∠OAC+90°=180°=>∠OAC=30°

Hence, we get position of C with respect to A is 20m west of south at angle of 30°