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$\frac{{{h_a}}}{{{l_a}}} = \frac{{AH}}{{AD}} = \cos \widehat {HAD}$
• Trường hợp $B \le {90^\circ}$ $\begin{array}{l} \widehat {HAD} = \widehat {BAD} - \widehat {BAH}\\ \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{A}{2} - \left( {{{90}^\circ} - B} \right)\\ \,\,\,\,\,\,\,\,\,\,\,\, = \frac{A}{2} - \left( {\frac{{A + B + C}}{2} - B} \right)\\ \,\,\,\,\,\,\,\,\,\,\, = \frac{{B - C}}{2} \end{array}$
• Trường hợp $B > {90^\circ}$ $\begin{array}{l} \widehat {HAD} = \widehat {BAD} + \widehat {BAH}\\ \,\,\,\,\,\,\,\,\,\,\,\, = \frac{A}{2} + \left( {B - {{90}^\circ}} \right)\\ \,\,\,\,\,\,\,\,\,\,\, = \frac{A}{2} + \left( {B - \frac{{A + B + C}}{2}} \right)\\ \,\,\,\,\,\,\,\,\,\,\, = \frac{{B - C}}{2} \end{array}$ Tóm lại trong cả hai trường hợp đều có $\widehat {HAD} = \frac{{B - C}}{2}$ . Vậy $\frac{{{h_a}}}{{{l_a}}} = \cos\widehat {HAD} = \cos\frac{{B - C}}{2}$ (đpcm).
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Đăng bài 04-05-12 05:45 PM
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