InterviewSolution
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InterviewSolution
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In the ambiguous case, if `a, b and A` are given and `c_1, c_2` are the two values of the third `(c_1-c_2)^2 + (c_1+c_2)^2 tan^2 A` is equal toA. 4B. `4a^(2)`C. `4b^(2)`D. `4c^(2)` |
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Answer» Correct Answer - B We have, `cosA=(b^(2)+c^(2)-a^(2))/(2bc)` `impliesc^(2)-(2bcosA)c+(b^(2)-a^(2))=0` Since `c_(1)` and `c_(2)` are the roots of this equation. `therefore c_(1)+c_(2)=2b cosA and c_(1)c_(2)=b^(2)-a^(2)` Now, `(c_(1)-c_(2))^(2)+(c_(1)+c_(2))^(2)tan^(2)A` `=(c_(1)+c_(2))^(2)-4c_(1)c_(2)+(c_(1)+c_(2))^(2)tan^(2)A` `=(c_(1)+c_(2))^(2)sec^(2)A-4c_(1)c_(2)` `=4b^(2)cos^(2)Axxsec^(2)A-4(b^(2)-a^(2))=4a^(2)` |
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