A is 40 , m South-west of B. C is 40 , m Sout | vedclass

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(A) Let $B$ be at the origin $(0,0)$ of the coordinate system.Since $A$ is $40 \, m$ South-west of $B$,its position can be represented as $A = (-40 \s

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East

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(A) Let $B$ be at the origin $(0,0)$ of the coordinate system.Since $A$ is $40 \, m$ South-west of $B$,its position can be represented as $A = (-40 \sin 45^{\circ}, -40 \cos 45^{\circ}) = (-20\sqrt{2}, -20\sqrt{2}).$Since $C$ is $40 \, m$ South-east of $B$,its position can be represented as $C = (40 \sin 45^{\circ}, -40 \cos 45^{\circ}) = (20\sqrt{2}, -20\sqrt{2}).$To find the direction of $C$ with respect to $A$,we look at the vector $\vec{AC} = C - A = (20\sqrt{2} - (-20\sqrt{2}), -20\sqrt{2} - (-20\sqrt{2})) = (40\sqrt{2}, 0).$Since the $y$-coordinate difference is $0$ and the $x$-coordinate difference is positive,$C$ lies directly to the East of $A$.

$A$ is $40 \, m$ South-west of $B.$ $C$ is $40 \, m$ South-east of $B.$ Then,$C$ is in which direction of $A$?

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