 Solution

Use $T=P_{V^{\perp }}{|}_U:U⟶V^{\perp }$. Positivity of the directed subspace angle cosine givesso $T$ is injective. The two vector spaces have the same finite dimension, $n$. By the rank-nullity theorem, $T$ is also surjective. There is no need to assume a second positive directed subspace angle cosine in this finite-dimensional case.

For any $f\in H$, find $u\in U$ with $Tu=P_{V^{\perp }}f$. Then $f-u\in V$. If $u\in U\cap V$, then $Tu=0$, and injectivity gives $u=0$. HenceThe direct sum is again bounded: $u=T^{-1}{P}_{V^{\perp }}f$. If $n=0$, then $U=\{0\}$ and $V^{\perp }=\{0\}$, so closedness of $V$ gives $V=H$ and the conclusion directly; no angle of an empty unit sphere is needed. Equal finite dimensions are essential to the surjectivity argument, whereas mere injectivity between infinite-dimensional Hilbert spaces is insufficient.