Answer:
The
angle which the given velocity makes with the north direction is\[\text{18}0{}^\circ
-\text{3}0{}^\circ =\text{15}0{}^\circ \]. The component velocity along north
\[=\text{v cos15}0{}^\circ =\text{1}00\times
(-\sqrt{3}/\text{2})=-\text{86}.\text{6 km}{{\text{h}}^{-\text{1}}}\].
The angle which the given velocity makes with the east
direction is\[\text{9}0{}^\circ +\text{3}0{}^\circ =\text{12}0{}^\circ \], The
component velocity along east
\[=\text{vcos 12}0{}^\circ =\text{1}00\times \left(
-\text{1}/\text{2} \right)=-\text{5}0\text{ km}{{\text{h}}^{-\text{1}}}\].
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