Mathematics - Circle Question with Solution | TestHub

$\begin{array}{rl}& \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}}\dots ..(1)\\ & \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{a}}+\beta \overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{a}}\\ & 0=\alpha +\beta \cos 15^{\circ }\dots .(2)\\ & (1)\Rightarrow \overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{c}}=\alpha \overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{b}}+\beta \overrightarrow{\mathrm{b}}\cdot \overrightarrow{\mathrm{b}}\\ & \Rightarrow \cos 75^{\circ }=\alpha \cos 15^{\circ }+\beta \dots .(3)\\ & (2)\&(3)\Rightarrow \cos 75^{\circ }=-\beta {\cos }^{2}15^{\circ }+\beta \end{array}$

$\beta =\frac{\cos 75^{\circ }}{{\sin }^{2}15^{\circ }}=\frac{1}{\sin 15^{\circ }}=\frac{2\sqrt{2}}{\sqrt{3}-1}$
(2) $\Rightarrow \alpha =\frac{-\cos 15^{\circ }}{\sin 15^{\circ }}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}$
$\therefore \overrightarrow{\mathrm{c}}=\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}\overrightarrow{\mathrm{a}}+\left(\frac{2\sqrt{2}}{\sqrt{3}-1}\right)\overrightarrow{\mathrm{b}}$

Now
$\begin{array}{rl}& \alpha +\sqrt{2}(\sqrt{3}-1)\beta =\frac{-(\sqrt{3}+1)}{(\sqrt{3}-1)}+\frac{\sqrt{2}(\sqrt{3}-1)\cdot 2\sqrt{2}}{\sqrt{3}-1}\\ & =\frac{-(\sqrt{3}+1{)}^2}{2}+4\\ & =\frac{-3-1-2\sqrt{3}+8}{2}\\ & =2-\sqrt{3}\end{array}$