Given,
Magnitude of the vector = 40 units
Angle =45∘= 45^\circ=45∘ North of east East
Let the vector
Horizontal component (rx→)=40cos(45∘)i^(\overrightarrow{r_x})=40 \cos (45^\circ) \hat{i}(rx)=40cos(45∘)i^
Vertical component (ry→)=40sin(45∘)j^(\overrightarrow{r_y})=40\sin(45^\circ)\hat{j}(ry)=40sin(45∘)j^
Hence the required vector r→=40cos(45∘)i^+40sin(45∘)j^\overrightarrow{r}=40\cos (45^\circ)\hat{i}+40\sin(45^\circ)\hat{j}r=40cos(45∘)i^+40sin(45∘)j^
⇒∣r∣=(40cos(45∘))2+(40sin(45∘))2\Rightarrow |r|=\sqrt{(40\cos(45^\circ))^2+(40\sin (45^\circ))^2}⇒∣r∣=(40cos(45∘))2+(40sin(45∘))2
⇒∣r∣=202+202=202\Rightarrow |r|=\sqrt{20^2+20^2}=20\sqrt{2}⇒∣r∣=202+202=202
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