解: ∵$\angle A=58°， $∴$\angle A B C+\angle A C B=180°-\angle A=18 0°-58°=122° ①，$
∵${BH} $是$ \angle {ABC} $的平分线, ∴$\angle H B C=\frac {1}{2} \angle {ABC} ，$
∵$\angle A C D $是$ \triangle A B C $的外角$， C H $是外角$ \angle A C D $的角平分线,
∴$\angle {ACH}=\frac {1}{2}(\angle {A}+\angle {ABC})， $
∴$\angle {BCH}=\angle {ACB}+\angle {ACH}=\angle {ACB}+\frac {1}{2}(\angle {A}+\angle {AB}C)，$
∵$\angle {H}+\angle {HBC}+\angle {ACB}+\angle {ACH}=180°， $
∴$\angle {H}+\frac {1}{2} \angle {ABC}+\angle {ACB}+\frac {1}{2}(\angle {A}+\angle {ABC})=180°，$
$ 即 \angle {H}+(\angle {ABC}+\angle {ACB})+\frac {1}{2} \angle {A}=180°②， $
$ 把①代入②得， \angle {H}+122°+\frac {1}{2} ×58°=180°， $
∴$\angle H=29° .$
