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🅿🅰🆆🅰🅽 . 7 years, 10 months ago
= -cos (1710)°
=-cos (180*10-90)°
= -cos (-90)°
= cos (90)°
= 0
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Rasmi Rv 8 years, 4 months ago
{tex}Let\; the \quad angles\ be\quad x,2x\; and\; 3x\quad then,x+2x+3x={ 180^{ o } }\Rightarrow 6x=180\Rightarrow x{ { =\frac { 180 }{ 6 } } }={ 30^{ o\\ } }\\ Using\; sine\quad rule\quad we\quad have\frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } =k(say)\\ Herea=k\sin 30,b=k\sin 60,c=k\sin 90\\ \Rightarrow a=\frac { { { k } } }{ 2 } ,b={ { \frac { \sqrt { 3 } }{ 2 } } }.k,c={ 1 }.k\\ a{ { :b{ { : } }c } }=\frac { { { k } } }{ 2 } { { :\frac { k\sqrt { 3 } }{ 2 } { { : } } } }{ 1 }.k\\ \Rightarrow a{ { :b{ { : } }c } }=1{ { :\sqrt { 3 } { { : } }2 } }{/tex}
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Raveesh Yadav 8 years, 4 months ago
With teh help of analyser
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