To find :Angle between vectors [tex]\sf{\vec{A}}[/tex] = 5 [tex]\sf{\hat{i}}[/tex] + [tex]\sf{\hat{j}}[/tex]  and [tex]\sf{\vec{B}}[/tex] = 2 [tex]\sf{\hat{i}}[/tex] + 4 [tex]\sf{\hat{j}}[/tex] Answer : Dot product of [tex]\sf{\vec{A}}[/tex] and [tex]\sf{\vec{B}}[/tex][tex]:\implies\sf{\vec{A}\;.\;\vec{B}= (5\;\hat{i} + \hat{j} ) \;.\; ( 2\;\hat{i} + 4\;\hat{j})}[/tex][tex]:\implies\sf{\vec{A}\;.\;\vec{B}= (5) (2) + (1) (4) }[/tex][tex]:\implies\sf{\vec{A}\;.\;\vec{B}= 14}[/tex] Magnitude of vectors [tex]:\implies\sf{| \vec{A} | = \sqrt{ (5)^2 + (1)^2 } }[/tex][tex]:\implies\sf{| \vec{A} | = \sqrt{26} }[/tex]and[tex]:\implies\sf{|B|=\sqrt{(2)^2+ (4)^2}}[/tex][tex]:\implies\sf{|B|=\sqrt{20}}[/tex] Let, Angle between vectors [tex]\sf{\vec{A}}[/tex] and [tex]\sf{\vec{B}}[/tex] be [tex]\theta[/tex]then,[tex]:\implies\sf{cos\;\theta=\dfrac{\vec{A} \;.\;\vec{B}}{|\vec{A}|\;.\;|\vec{B}|}}[/tex][tex]:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{26}\;.\;\sqrt{20}}}[/tex][tex]:\implies\sf{cos\;\theta=\dfrac{14}{\sqrt{520}}}[/tex][tex]:\implies\sf{cos\;\theta=\dfrac{14}{2\sqrt{130}}}[/tex][tex]:\implies\sf{cos\;\theta=\dfrac{7}{\sqrt{130}}}[/tex][tex]:\implies\sf{cos\;\theta=\dfrac{7\sqrt{130}}{130}}[/tex][tex]:\implies\sf{\theta=cos^{-1}\bigg(\dfrac{7\sqrt{130}}{130}\bigg)}[/tex][tex]:\implies\boxed{\boxed{\large{\sf{\;\;\theta=52.125\;^{\circ}\;}}}}[/tex]Therefore,Angle between vectors [tex]\sf{\vec{A}}[/tex] and [tex]\sf{\vec{B}}[/tex] will be 52.125°. (Approx.) Answer from Cosmique on brainly.in