△ABC -এর ∠C=90\(^o\), যদি BC = \(m\) এবং AC = \(n\) হয় তবে দেখাও যে, \(msinA+nsinB\) \(=\sqrt{m^2+n^2}\) Madhyamik 2019


\( AB=\sqrt{BC^2+AC^2}=\sqrt{m^2+n^2}\)
\(\therefore msinA+nsinB=m.\cfrac{BC}{AB}+n.\cfrac{AC}{AB}\)
\(=m.\cfrac{m}{\sqrt{m^2+n^2}}+n.\cfrac{n}{\sqrt{m^2+n^2}}\)
\(=\cfrac{m^2}{\sqrt{m^2+n^2}}+\cfrac{n^2}{\sqrt{m^2+n^2}}\)
\(=\cfrac{m^2+n^2}{\sqrt{m^2+n^2}}\)
\(=\sqrt{m^2+n^2}\) (প্রমাণিত)

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