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| 1. |
In an acute angled triangle ABC, if tan(A+B-C) =1 and ,sec(B+C-A) =2,find the value of A, B andC |
| Answer» According to the question, tan (A+B-C) = 1\xa0{tex}\\Rightarrow{/tex}\xa0tan (A+B-C) =\xa0{tex}\\tan 45 ^ { \\circ }{/tex}{tex}\\Rightarrow{/tex}\xa0A + B - C = 45° ........(1)Also given, sec (B+C-A) = 2{tex}\\Rightarrow{/tex}\xa0sec (B + C - A ) = sec 60°\xa0{tex}\\therefore{/tex}{tex}B + C - A = 60 ^ { \\circ }{/tex}...........(2)Adding equation (1) & (2);{tex}( A + B - C ) + ( B + C - A ) = 45 ^ { \\circ } + 60 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad 2 B = 105 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad B = 52 \\frac { 1 ^ { \\circ } } { 2 }{/tex}Putting\xa0{tex}B = 52 \\frac { 1 ^ { \\circ } } { 2 }{/tex}\xa0in equation (2); we get :-{tex}52 \\frac { 1 ^ { \\circ } } { 2 } + C - A = 60 ^ { \\circ }{/tex}{tex}\\Rightarrow C - A = 7 \\frac { 1 ^ { \\circ} } { 2 }{/tex}......(3)Also, in\xa0{tex}\\Delta A B C{/tex}, we have\xa0{tex}A + B + C = 180 ^ { \\circ }{/tex}{tex}\\Rightarrow \\quad A + 52 \\frac { 1 ^ { \\circ } } { 2 } + C = 180 ^ { \\circ }{/tex}\xa0{tex}\\left[ \\because B = 52 \\frac { 1 ^ { \\circ } } { 2 } \\right]{/tex}{tex}\\Rightarrow \\quad C + A = 127 \\frac { 1 ^ { \\circ } } { 2 }{/tex}......(4)Adding and subtracting (3) and (4), we get\xa02C = 135o and 2A = 120o{tex}\\Rightarrow C = 67 \\frac { 1 } { 2 } ^ { \\circ } {/tex}\xa0and A = 60oHence, we get the values of A = 60o,{tex}B = 52 \\frac { 1 ^ { \\circ } } { 2 }{/tex}and\xa0{tex}C = 67 \\frac { 1 ^ { \\circ } } { 2 }{/tex}.\xa0 | |