question_answer

Each side of an equilateral triangle subtends an angle of \[60{}^\circ \]at the top of a tower h m high located at the centre of the triangle. If a is the length of each of side of the triangle, then

A) \[3{{a}^{2}}=2{{h}^{2}}\]

B) \[2{{a}^{2}}=3{{h}^{2}}\]

C) \[{{a}^{2}}=3{{h}^{2}}\]

D) \[3{{a}^{2}}={{h}^{2}}\]

Correct Answer: B

Solution :

[b] Let QT be the tower of height (h) I \[\Delta PRS.\]now, each triangle QPR, QRS, QSP ar equilateral.

Thus QP=QS=QR=a.

In \[\Delta QTP,\]

\[Q{{P}^{2}}=Q{{T}^{2}}+P{{T}^{2}}\]

\[\Rightarrow {{a}^{2}}={{h}^{2}}+{{\left( \frac{a}{2}\sec 30{}^\circ  \right)}^{2}}\]

\[\Rightarrow {{a}^{2}}={{h}^{2}}+\frac{{{a}^{2}}}{4}.\frac{4}{3}\]

\[\Rightarrow {{a}^{2}}={{h}^{2}}+\frac{{{a}^{2}}}{3}\]

\[\Rightarrow {{a}^{2}}-\frac{{{a}^{2}}}{3}={{h}^{2}}\]

\[\Rightarrow \frac{3{{a}^{2}}-{{a}^{2}}}{3}={{h}^{2}}\therefore 2{{a}^{2}}=3{{h}^{2}}\]