A) \[\frac{2\sqrt{{{b}^{2}}-{{a}^{2}}}}{4{{b}^{2}}-3{{a}^{2}}}\]
B) \[\frac{a}{\sqrt{4{{b}^{2}}-3{{a}^{2}}}}\]
C) \[\frac{\sqrt{4{{b}^{2}}-3{{a}^{2}}}}{2}\]
D) None of these
Correct Answer: A
Solution :
| \[AD=DB=\frac{a}{2}\] |
| Also, \[B{{C}^{2}}+A{{B}^{2}}=A{{C}^{2}}\](by Pythagoras theorem) |
| In \[\Delta ABC,\] |
| \[\Rightarrow \] \[B{{C}^{2}}+{{a}^{2}}={{b}^{2}}\] |
| \[\Rightarrow \] \[B{{C}^{2}}={{b}^{2}}-{{a}^{2}}\] |
| \[\Rightarrow \] \[BC=\sqrt{{{b}^{2}}-{{a}^{2}}}\] |
| In \[\Delta BCD,\] \[C{{D}^{2}}=B{{C}^{2}}+B{{D}^{2}}\] |
| \[C{{D}^{2}}={{(\sqrt{{{b}^{2}}-{{a}^{2}}})}^{2}}+{{\left( \frac{a}{2} \right)}^{2}}=\left( {{b}^{2}}-{{a}^{2}}+\frac{{{a}^{2}}}{4} \right)\] |
| \[C{{D}^{2}}=\frac{4{{b}^{2}}-3{{a}^{2}}}{4}\] |
| \[CD=\frac{\sqrt{4{{b}^{2}}-3{{a}^{2}}}}{2}\] |
| \[\therefore \]\[\cos \theta =\frac{BC}{CD}=\frac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{\sqrt{\frac{4{{b}^{2}}-3{{a}^{2}}}{2}}}=\frac{2\sqrt{{{b}^{2}}-{{a}^{2}}}}{\sqrt{4{{b}^{2}}-3{{a}^{2}}}}\] |
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