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Find cosec30° and cos60° geometrically |
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Answer» Let a triangle ABC with each side equal to 2a{tex}\\angle{/tex}A = {tex}\\angle{/tex}B = {tex}\\angle{/tex}C = 60°Draw AD perpendicular to BCIn\xa0{tex}\\triangle{/tex}BDA and\xa0{tex}\\triangle{/tex}CDAAB = AC, AD = AD and\xa0{tex}\\angle BDA=\\angle CDA{/tex}{tex}\\triangle B D A \\cong \\triangle C D A{/tex}\xa0by RHSBD = CD{tex}\\angle{/tex}BAD = {tex}\\angle{/tex}CAD = 30° by CPCTIn {tex}\\triangle{/tex}BDA,\xa0{tex}\\text{cosec} 30 ^ { \\circ } =\\frac HP= \\frac { A B } { B D } = \\frac { 2 a } { a } = 2{/tex}\xa0and\xa0{tex}\\cos 60 ^ { \\circ } =\\frac BH= \\frac { B D } { A B } = \\frac { a } { 2 a } = \\frac { 1 } { 2 }{/tex} let a triangle ABC with each side equal to 2a/_ |
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