A geometry problem by Mayank Srivastava

$\begin{aligned} \cot {70^\circ} + 4\cos {70^\circ} & = \sqrt{k} \\ \dfrac {\cos {70^\circ}}{\sin {70^\circ}} + 4\cos {70^\circ} & = \sqrt{k} \\ \cos {70^\circ} + 4\sin {70^\circ} \cos {70^\circ} & = \sqrt{k} \sin {70^\circ} \\ \cos {70^\circ} + 2\sin {140^\circ} & = \sqrt{k} \sin {70^\circ} \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = - 2\sin {140^\circ} \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = 2 \cos {130^\circ} \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = 2 ( \cos {60^\circ} \cos {70^\circ} - \sin {60^\circ} \sin {70^\circ} ) \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = 2 \left( \dfrac{1}{2} \cos {70^\circ} - \dfrac{\sqrt{3}}{2} \sin {70^\circ} \right) \\ \cos {70^\circ} - \sqrt{k} \sin {70^\circ} & = \cos {70^\circ} - \sqrt{3} \sin {70^\circ} \\ \implies k & = \boxed{3} \end{aligned}$

Prepare for a trig identity overload. Most of this will deal with what is inside the parentheses, and will be squared at the very end. All of this is performed in degrees.

Trig Identity 1: $\cot\theta = \frac{\cos\theta}{\sin\theta}$

$\cot70 + 4\cos70 = \frac{\cos70}{\sin70} + 4\cos70 = \frac{\cos70 + 4\sin70 \cos70}{\sin70}$

Trig Identity 2: $\sin2\theta = 2\sin\theta\cos\theta$

Trig Identity 3: $\cos(90-\theta) = \sin\theta$

$=\frac{\cos(90-20) + 2(2\sin70\cos70)}{\sin70} = \frac{\sin20 + 2\sin140}{\sin70}$ $= \frac{\sin20 + \sin140 + \sin140}{\sin70}$

Trig Identity 4: $\sin\theta+\sin\phi=2\sin(\frac{\theta+\phi}{2})\cos(\frac{\theta-\phi}{2})$

Unit Circle: $\cos60=\frac{1}{2}$

Unit Circle: $\cos30=\frac{\sqrt{3}}{2}$

$=\frac{2\sin80\cos60+\sin140}{\sin70}=\frac{\sin80+\sin140}{\sin70}$

$=\frac{2\sin110\cos30}{\sin70}=\frac{\sqrt{3}\sin110}{\sin70}$

Trig Identity 5: $\sin(180-\theta)=\sin\theta$

$=\frac{\sqrt{3}\sin(180-70)}{\sin70}=\frac{\sqrt{3}\sin70}{\sin70}=\sqrt{3}$

This was all squared, therefore $(\sqrt{3})^{2}=3\space\space\space\Box$