mathop {,{rm{A}}}limits^ to ,, + ;,mathop {rm | vedclass

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Question

$\begin{array}{l}
\mathop {\,{m{A}}}\limits^ 	o  \,\, + \;\,\mathop {m{B}}\limits^ 	o  \,\, + \,\,\mathop {m{C}}\limits^ 	o

Vedclass · Accepted Answer

$90^&deg;,135^&deg;,135^&deg;$

Vedclass · Answer

$\begin{array}{l}
\mathop {\,{m{A}}}\limits^ 	o  \,\, + \;\,\mathop {m{B}}\limits^ 	o  \,\, + \,\,\mathop {m{C}}\limits^ 	o  \, = \,\,\mathop 0\limits^ 	o  \, \Rightarrow \,\,\mathop {\,{m{A}}}\limits^ 	o  \,\, + \;\,\mathop {m{B}}\limits^ 	o  \,\, =  - \,\,\mathop {m{C}}\limits^ 	o  \
{A^2}\,\, + \;\,{B^2}\,\, + \;\,2AB\cos {	heta _1}\,\, = \,\,{C^2}\,\,\left( {\,\,A\,\, = \,B} ight)\
{B^2}\,\, + \;\,{B^2}\,\, + \;\,2AB\cos {	heta _1}\,\, = \,\,2{B^2}\,\,\left( {\,\,C\,\, = \,\,\sqrt 2 B} ight)\
2AB\cos {	heta _1}\,\, = \,\,0\,\, \Rightarrow \,\,\cos {	heta _1}\,\, = \,\,0\,\, \Rightarrow \,\,{	heta _1}\,\, = \,\,90^\circ \
\mathop {m{B}}\limits^ 	o  \,\, + \,\,\mathop {m{C}}\limits^ 	o  \,\, = \,\, - \mathop {m{A}}\limits^ 	o  \,\, \Rightarrow \,\,{B^2}\,\, + \;\,{C^2}\,\, + \;\,2BC\,\,\cos {	heta _2}\,\, = \,\,{A^2}\
{B^2}\,\, + \;2{B^2}\,\, + \;\,2B\sqrt 2 \,B\cos {	heta _2}\,\, = \,\,{B^2}\
2{B^2}\,\,\left( {1\, + \,\,\sqrt 2 \cos {	heta _2}} ight)\,\, = \,\,0\,\, \Rightarrow \,\,\cos {	heta _2}\,\, = \,\, - \frac{1}{{\sqrt 2 }}\,\, \Rightarrow \,\,{	heta _2}\,\, = \,\,135^\circ \
	herefore \,\,\left( {90^\circ ,\,135^\circ ,135^\circ } ight)
\end{array}$ ખૂણાઓ છે

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