Find the magnitude and direction angle (to the nearest tenth) for each vector. Give the measure of the direction angle as
an angle in [0,360°].
-$\langle 8 \sqrt { 3 } , 8 \rangle$
A) $16 \sqrt { 2 } ; 60 ^ { \circ }$
B) $32 ; 30 ^ { \circ }$
C) $16 ; 30 ^ { \circ }$
D) $16 ; 150 ^ { \circ }$

Question

Quizplus · Accepted Answer

The magnitude is calculated as $\sqrt{(8\sqrt{3})^2 + 8^2} = \sqrt{192 + 64} = \sqrt{256} = 16$. The direction angle is $\tan^{-1}\left(\frac{8}{8\sqrt{3}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = 30^\circ$.

Find the Magnitude and Direction Angle (To the Nearest Tenth) ⟨83,8⟩\langle 8 \sqrt { 3 } , 8 \rangle⟨83​,8⟩

Find the Magnitude and Direction Angle (To the Nearest Tenth) $\langle 8 \sqrt { 3 } , 8 \rangle$