In the figure AD=4cm BD=3cm CB=12cm then find the value of cot theta
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In right\xa0{tex}\\triangle{/tex}ABD,\xa0{tex}\\angle D = 90 ^ { \\circ }{/tex}{tex}AB^2 = AD^2 + BD^2{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}AB^2 = 4^2 + 3^2{/tex}{tex}\\Rightarrow{/tex}\xa0AB =\xa0{tex}\\pm 5{/tex}\xa0cm{tex}\\Rightarrow{/tex}\xa0{tex}AB = 5 cm{/tex} (Neglecting negative value)In right\xa0{tex}\\triangle{/tex}ABC , cot\xa0{tex}\\theta{/tex}\xa0=\xa0{tex}\\frac { \\mathrm { BC } } { \\mathrm { AB } } = \\frac { 12 } { 5 }{/tex}