Câu hỏi:

Có \(\widehat {CAH} = {\alpha _1} = 30{^ \circ },\,\,\widehat {CBH} = {\beta _1} = 70{^ \circ } \Rightarrow \widehat {ACB} = \widehat {CBH} - \widehat {CAH} = 40{^ \circ }\).

Áp dụng định lí sin vào \({\rm{\Delta }}ABC\), ta có: \(\frac{{BC}}{{{\rm{sin}}\widehat {CAH}}} = \frac{{AB}}{{{\rm{sin}}\widehat {ACB}}} \Rightarrow BC = \frac{{20{\rm{sin}}30{^ \circ }}}{{{\rm{sin}}40{^ \circ }}}\)

Xét \({\rm{\Delta }}HBC\) vuông tại H, ta có: \({\rm{sin}}\widehat {CBH} = \frac{{CH}}{{BC}} \Rightarrow CH = BC{\rm{sin}}\widehat {CBH} = \frac{{20{\rm{sin}}30{^ \circ }}}{{{\rm{sin}}40{^ \circ }}} \cdot {\rm{sin}}70{^ \circ }\).

Có \(\widehat {OAH} = {\alpha _2} = 50{^ \circ },\,\,\widehat {OBH} = {\beta _2} = 80{^ \circ } \Rightarrow \widehat {AOB} = 30{^ \circ }\).

Áp dụng định lí sin vào \({\rm{\Delta }}ABO\), ta có: \(\frac{{BO}}{{{\rm{sin}}\widehat {OAH}}} = \frac{{AB}}{{{\rm{sin}}\widehat {AOB}}} \Rightarrow BO = \frac{{20{\rm{sin}}50{^ \circ }}}{{{\rm{sin}}30{^ \circ }}}\).

Xét \({\rm{\Delta }}HBO\) vuông tại H, ta có: \({\rm{sin}}\widehat {OBH} = \frac{{HO}}{{BO}} \Rightarrow HO = BO{\rm{sin}}\widehat {OBH} = \frac{{20{\rm{sin}}50{^ \circ }}}{{{\rm{sin}}30{^ \circ }}} \cdot {\rm{sin}}80{^ \circ }\).

Vậy  \(h = OC = HO - CH = \frac{{20{\rm{sin}}50{^ \circ }}}{{{\rm{sin}}30{^ \circ }}} \cdot {\rm{sin}}80{^ \circ } - \frac{{20{\rm{sin}}30{^ \circ }}}{{{\rm{sin}}40{^ \circ }}} \cdot {\rm{sin}}70{^ \circ } \approx 15,56\)(m).