Math, asked by rrr2273, 1 year ago

in a right triangle ABC right angled at C in which AB= 13 cm BC = 5 cm determine the value of cos square B + sin square A​


Eshu6789: hey listen there was some mistake
Eshu6789: your correct ans is-
Eshu6789: 50/169
Eshu6789: bcz sin^2A=(AC/AB)^2
Eshu6789: By uising it calculate the full math...

Answers

Answered by MaheswariS
9

\underline{\textbf{Given:}}

\textsf{In right triangle ABC,}

\mathsf{\angle{C}=90^\circ,\;AB=13\;cm,\;BC=5\,cm}

\underline{\textbf{To find:}}

\textsf{The value of}

\mathsf{cos^2B+sin^2A}

\underline{\textbf{Solution:}}

\mathsf{In\;right\;\triangle\;ABC,}

\mathsf{sin\,A=\dfrac{BC}{AB}=\dfrac{5}{13}}

\mathsf{cos\,B=\dfrac{BC}{AB}=\dfrac{5}{13}}

\mathsf{Now,}

\mathsf{cos^2B+sin^2A}

\mathsf{=\left(\dfrac{5}{13}\right)^2+\left(\dfrac{5}{13}\right)^2}

\mathsf{=\dfrac{25}{169}+\dfrac{25}{169}}

\mathsf{=\dfrac{50}{169}}

\implies\boxed{\mathsf{cos^2B+sin^2A=\dfrac{50}{169}}}

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