question_answer

In a \[\Delta ABC\] if \[\angle A=30{}^\circ \]and \[BC=2+\sqrt{5},\] then the distance of the vertex A from the orthocentre of the \[\Delta ABC\] is

A) \[2\,(2+\sqrt{5})\]

B) \[\sqrt{3}\,(2+\sqrt{5})\]

C) \[\frac{\sqrt{5}+2}{2\sqrt{2}}\]

D) \[\frac{2+\sqrt{5}}{\sqrt{3}}\]

Correct Answer: B

Solution :

In \[\Delta ABC,\]\[\angle A=30{}^\circ \], \[BC=2+\sqrt{5}\]

\[\frac{a}{\sin A}=2R\]\[\Rightarrow \]\[\frac{2+\sqrt{5}}{\sin 30{}^\circ }=2R\]

\[R=\frac{2+\sqrt{5}}{2\times \frac{1}{2}}=2+\sqrt{5}\]

Now,     \[AH=2R\cos A\]

\[\because \]\[AH=2\,(2+\sqrt{5})\cos 30{}^\circ \]\[\Rightarrow \]\[AH=2\,(2+\sqrt{5})\frac{\sqrt{3}}{2}=(2+\sqrt{5})(\sqrt{3})\]

\[\therefore \]\[AH=\sqrt{3}\,(2+\sqrt{5})\]