InterviewSolution
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InterviewSolution
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\sin \theta=\frac{m^{2}+2 m n}{m^{2}+2 m n+2 n^{2}}\begin{array}{l}{\frac{m(m+2 n)}{2 n(m-n)}} \\ {\frac{m(m+2 n)}{2 n\left(m+n^{2}\right)}}\end{array} \quad(b) \frac{m(m-2 n)}{2 n(m+n)} |
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Answer» hypotaneous=m^2+2mn+n^2remaining one side=m^2+2mnso other side=xx^2=(m^2+2mn+n^2)^2-(m^2+2mn)^2 =m^4+4m^2n^2+n^4+4m^3n+4mn^3+2m^2n^2-m^4-4m^2n^2-4m^3n=n^4+4mn^3+2m^2n^2=n^2(m^2+2mn+n^2)=n^2(m+n)^2so x=n(m+n) so tanx=(m^2+2mn)/(n(m+n)) =(m(m+2n))/(n(m+n)) wro wrong |
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