= Solution
The sum of the degrees of all <face of a planar map>[faces] is $2e$. Each of the $t$ triangular faces contributes three. Because the graph is bridgeless and has no four-cycle, every other face has degree at least five, and therefore
$$
\boxed{2e\ge3t+5(f-t)}.
$$
No edge can border two triangular faces: two distinct triangles sharing that edge would have their other two edges form a four-cycle, while the same triangle on both sides would force the connected bridgeless graph to be that triangle, contrary to $n\ge4$. Thus the $3t$ edge incidences belonging to triangular faces use distinct edges, so
$$
\boxed{e\ge3t}.
$$
Combining the inequalities gives
$$
2e\ge5f-2t\ge5f-\frac{2e}{3},
$$
and hence
$$
\boxed{f\le\frac{8e}{15}}.
$$
The <Euler formula for a connected planar graph> now yields
$$
2=n-e+f
\le n-e+\frac{8e}{15}
=n-\frac{7e}{15},
$$
so the <triangle-pentagon planar edge bound> is
$$
\boxed{e\le\frac{15(n-2)}7}.
$$
Equality is possible. Cut each corner of a <dodecahedron> through the midpoints of its incident edges. The resulting <icosidodecahedral graph> has
$$
n=30,\qquad e=60,\qquad t=20,\qquad f-t=12,
$$
with every edge incident to one triangular and one pentagonal face. It has no four-cycle, and
$$
60=\frac{15(30-2)}7.
$$
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