Question

Given
( bar{A}+bar{B}=bar{C} )
Squaring on both sides we get
( bar{A}^{2}+bar{B}^{2}+2 bar{A} bar{B}=bar{C}^{2} )
( Rightarrow(12)^{2}+(5)^{2}+2 bar{A} bar{B}=(13)^{2} )
( Rightarrow 144+25+2 bar{A} bar{B}=169 )
( Rightarrow 169+2 bar{A} bar{B}=169 )
( Rightarrow 2 bar{A} bar{B}=0 )
( Rightarrow bar{A} bar{B}=0 )
Let be the angle between ( A ) and ( B ).
( cos theta=frac{bar{A} bar{B}}{bar{A}^{2}+bar{B}^{2}} )
( Rightarrow cos theta=0 )
( therefore ) Angle between ( A ) and ( B ) is ( 90^{0} )

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Solution