Quantitative Aptitude Ques 1824

Question: $ \Delta ABC $ is right angled triangle, where $ \angle ABC=90{}^\circ . $ If $ AC=2\sqrt{5} $ and $ AB-BC=2, $ then the value of $ {{\cos }^{2}}A-{{\cos }^{2}}C $ is

Options:

A) $ \frac{1}{\sqrt{5}} $

B) $ \sqrt{5} $

C) $ \frac{1}{2} $

D) $ \frac{3}{5} $

Show Answer

Answer:

Correct Answer: D

Solution:

  • Let $ BC=x, $ then $ AB=x+2 $ In $ \Delta ABC, $ $ {{(x+2)}^{2}}+x^{2}={{(2\sqrt{5})}^{2}} $

$ \Rightarrow $ $ x^{2}+4+4x+x^{2}=20 $

$ \Rightarrow $ $ 2x^{2}+4x-16=0 $

$ \Rightarrow $ $ x^{2}+2x-8=0 $

$ \Rightarrow $ $ (x-2)(x+4)=0 $

$ \Rightarrow $ $ x=2 $ So, $ AB=4, $ $ BC=2 $

$ \therefore $ $ {{\cos }^{2}}A-{{\cos }^{2}}C={{( \frac{4}{2\sqrt{5}} )}^{2}}-{{( \frac{2}{2\sqrt{5}} )}^{2}}=\frac{3}{5} $

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