Find the value of tan 60 geometrically
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Let\xa0{tex}\\triangle{/tex}ABC\xa0is an equilateral A with each side = 2a units. Draw\xa0{tex}\\mathrm { AD } \\perp \\mathrm { BC }{/tex}{tex}\\therefore{/tex}\xa0D is mid-point of BC{tex}\\Rightarrow{/tex}\xa0BD = aIn right\xa0{tex}\\triangle{/tex}ADBAB2 = BD2 + AD2{tex}\\Rightarrow{/tex}\xa0(2a)2 = a2 + AD2{tex}\\Rightarrow{/tex}\xa04a2 – a2 = AD2{tex}\\Rightarrow{/tex}\xa0AD =\xa0{tex}\\sqrt { 3 a ^ { 2 } } = \\sqrt { 3 } a{/tex}Now in right\xa0{tex}\\triangle{/tex}ADBtan B\xa0{tex}= \\frac { \\mathrm { AD } } { \\mathrm { BD } }{/tex}{tex}\\Rightarrow{/tex}\xa0tan\xa0{tex}60 ^ { \\circ } = \\frac { \\sqrt { 3 } a } { a }{/tex}\xa0{tex}\\left( \\because \\angle B = 60 ^ { \\circ } \\right){/tex}{tex}\\Rightarrow{/tex}\xa0tan\xa0{tex}60 ^ { \\circ } = \\sqrt { 3 }{/tex}\xa0