The distance between vertex and orthocentre of a triangle

In $ \displaystyle \Delta \ ABC,\ \angle A={{60}^{\circ }},\ BC=5$ ,then find the distance of the vertex A from the orthocentre of  $ \displaystyle \Delta \ ABC\ $ .

Solution

$ \displaystyle BF=5\sin \theta \ \ \ ,\ \ CF=5\cos \theta $ 

$ \displaystyle GF=5\sin \theta \tan {{30}^{\circ }}$

$ \displaystyle \ \ \ \ \ \ =\frac{5}{{\sqrt{3}}}\sin \theta $ 

$ \displaystyle \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{sines}\ ,$ 

$ \displaystyle \frac{{AG}}{{\sin {{{30}}^{\circ }}}}=\frac{{CG}}{{\sin \ ({{{60}}^{\circ }}-\theta )}}$ 

$ \displaystyle \ \ \ 2AG=\frac{{CF-GF}}{{\sin {{{60}}^{\circ }}\cos \theta -\cos {{{60}}^{\circ }}\sin \theta }}$

$ \displaystyle \ \ \ 2AG=\frac{{5\cos \theta -\frac{5}{{\sqrt{3}}}\sin \theta }}{{\frac{{\sqrt{3}}}{2}\cos \theta -\frac{1}{2}\sin \theta }}$

$ \displaystyle \ \ \ \ \ AG=\frac{{\frac{5}{{\sqrt{3}}}(\sqrt{3}\cos \theta -\sin \theta )}}{{\sqrt{3}\cos \theta -\sin \theta }}$

$ \displaystyle \ \ \therefore \ AG=\frac{5}{{\sqrt{3}}}$