In △ABC\triangle ABC△ABC, HHH is the orthocenter. If ∠BAC=70∘\angle BAC = 70^\circ∠BAC=70∘, find ∠BHC\angle BHC∠BHC.
70∘70^\circ70∘
110∘110^\circ110∘
140∘140^\circ140∘
125∘125^\circ125∘