Trigo 1

It appears that there is no algebraic solution to this so I am solving it numerically.

From $\cos 2\theta = 1 - 2\sin^2 \theta$ :

$\begin{aligned} 1-2\sin^2 15^\circ & = \cos 30^\circ \\ \implies \sin 15^\circ & = \sqrt{\frac {1- \cos 30^\circ}2} = \sqrt {\frac {2-\sqrt 3}4} = \frac 12 \sqrt{\frac {4-2\sqrt 3}2} = \frac 12 \sqrt{\frac {(\sqrt 3-1)^2}2} = \frac {\sqrt 3-1}{2\sqrt 2} \approx 0.258819045 \end{aligned}$

From $\tan 3 \theta = \dfrac {3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta}$ :

$\begin{aligned} \frac {3\tan 10^\circ - \tan^3 10^\circ}{1-3\tan^2 10^\circ} & = \frac 1{\sqrt 3} \\ \sqrt 3 \tan^3 10^\circ - 3 \tan^2 10^\circ - 3\sqrt 3 \tan 10^\circ + 1 & = 0 \end{aligned}$

Using Newton Raphson method , we get $\tan 10^\circ \approx 0.176326981$ .

Therefore, $5\sin 15^\circ + 5\tan 10^\circ \approx 5(0.258819045 + 0.176326981) \approx \boxed{2.176}$ .