Question

in right angled triangle ABC, angle A = 90° , AB = 8cm, BC = 17cm. find Sin B , Cos B . also prove that tan B × (90 - B)° = 1​​

Answer 1

$\textbf{Given:}$

$\mathsf{In\,\triangle\,ABC,\;\angle{A}=90^\circ,\;AB=8\,cm,\;BC=17\,cm}$

$\textbf{To find:}$

$\textsf{sinB, cosB}$

$\textbf{To prove:}$

$\mathsf{tanB{\times}tan(90^\circ-B)=1}$

$\textbf{Solution:}$

$\textsf{By Pythagoras theorem,}$

$\mathsf{BC^2=AB^2+AC^2}$

$\mathsf{17^2=8^2+AC^2}$

$\mathsf{289=64+AC^2}$

$\mathsf{AC^2=289-64}$

$\mathsf{AC^2=225}$

$\mathsf{AC=\sqrt{225}}$

$\implies\boxed{\mathsf{AC=15\,cm}}$

$\mathsf{sin\,B=\dfrac{AC}{BC}=\dfrac{15}{17}}$

$\mathsf{cos\,B=\dfrac{AB}{BC}=\dfrac{8}{17}}$

$\mathsf{tan\,B=\dfrac{AC}{AB}=\dfrac{15}{8}}$

$\mathsf{Here,\;C=90^\circ-B}$

$\mathsf{tan(90^\circ-B)=tan\,C=\dfrac{AB}{AC}=\dfrac{8}{15}}$

$\mathsf{Now,}$

$\mathsf{tanB{\times}tan(90^\circ-B)}$

$\mathsf{=\dfrac{15}{8}{\times}\dfrac{8}{15}}$

$\mathsf{=1}$

$\implies\boxed{\mathsf{tanB{\times}tan(90^\circ-B)=1}}$

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