Math, asked by praveenpandit1000, 12 hours ago

in figure,AC = 3cm ,BC = 6cm and CD = 4cm .Then sin A + cos B​

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Answered by MaheswariS
1

\underline{\textbf{Given:}}

\textsf{In figure, AC=3 cm, BC=6 cm}

\textsf{and CD=4 cm}

\underline{\textbf{To find:}}

\textsf{sinA+cosB}

\underline{\textbf{Solution:}}

\mathsf{In\;\triangle\,ACD,}

\mathsf{AD^2=AC^2+CD^2}

\mathsf{AD^2=3^2+4^2}

\mathsf{AD^2=9+16}

\mathsf{AD^2=25}

\mathsf{AD=\sqrt{25}}

\implies\boxed{\mathsf{AD=5\;cm}}

\mathsf{In\;\triangle\,BCD,}

\mathsf{BD^2=BC^2+CD^2}

\mathsf{BD^2=6^2+4^2}

\mathsf{BD^2=36+16}

\mathsf{BD^2=52}

\mathsf{BD=\sqrt{52}}

\implies\boxed{\mathsf{BD=2\sqrt{13}\;cm}}

\mathsf{Now,}

\mathsf{sinA+cosB}

\mathsf{=\dfrac{CD}{AD}+\dfrac{BC}{BD}}

\mathsf{=\dfrac{4}{5}+\dfrac{6}{2\sqrt{13}}}

\mathsf{=\dfrac{4}{5}+\dfrac{3}{\sqrt{13}}}

\mathsf{=\dfrac{4\sqrt{13}+15}{5\sqrt{13}}}

\underline{\textbf{Find more:}}

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