= Solution
Since $z>0$, the cone has the parametrization
$$
\phi(r,\theta)
=(r\cos\theta,r\sin\theta,\sqrt a\,r),
\qquad r>0.
$$
Its coordinate derivatives satisfy
$$
\phi_r\cdot\phi_r=1+a,
\qquad
\phi_r\cdot\phi_\theta=0,
\qquad
\phi_\theta\cdot\phi_\theta=r^2,
$$
so its <first fundamental form> is
$$
I=(1+a)\,dr^2+r^2\,d\theta^2.
$$
On a sufficiently narrow angular patch, define
$$
R=\sqrt{1+a}\,r,
\qquad
\Theta=\frac{\theta}{\sqrt{1+a}},
$$
and map the cone to the plane point with <polar coordinates> $(R,\Theta)$. The Euclidean metric pulls back as
$$
dR^2+R^2d\Theta^2
=(1+a)\,dr^2+r^2\,d\theta^2=I.
$$
The map is therefore a <local isometry from a circular cone to the plane>, proving that $\Sigma$ is locally isometric to the Euclidean plane.
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