a) ta có \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\Rightarrow\frac{a}{\sin A}=\frac{b+c}{\sin B+\sin C}=\frac{2a}{\sin B+\sin C}\)

do đó \(2a\cdot\sin A=a\left(\sin B+\sin C\right)\)

\(\Rightarrow2\sin A=\sin B+\sin C\)

b) ta có \(\frac{2}{h_a}=\frac{2a}{h_a\cdot a}=\frac{2a}{2S_{ABC}}=\frac{a}{S_{ABC}}\left(1\right)\)

\(\frac{1}{h_b}+\frac{1}{h_c}=\frac{b}{h_b\cdot b}+\frac{c}{h_c\cdot c}=\frac{b}{2S_{ABC}}+\frac{c}{2S_{ABC}}=\frac{b+c}{2S_{ABC}}=\frac{2a}{2S_{ABC}}=\frac{a}{S_{ABC}}\left(2\right)\)

từ (1) và (2) \(\Rightarrow\frac{2}{h_a}=\frac{1}{h_b}+\frac{1}{h_c}\)

Bài 14.

Áp dụng định lí hàm số Cô sin, ta có:

\(\dfrac{{{\mathop{\rm tanA}\nolimits} }}{{\tan B}} = \dfrac{{\sin A.\cos B}}{{\cos A.\sin B}} = \dfrac{{\dfrac{a}{{2R}}.\dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}}}}{{\dfrac{b}{{2R}}.\dfrac{{{c^2} + {b^2} - {a^2}}}{{2bc}}}} = \dfrac{{{c^2} + {a^2} - {b^2}}}{{{c^2} + {b^2} - {a^2}}} \)

Bài 19.

Áp dụng định lí sin và định lí Cô sin, ta có:

\( \cot A + \cot B + \cot C\\ = \dfrac{{R\left( {{b^2} + {c^2} - {a^2}} \right)}}{{abc}} + \dfrac{{R\left( {{c^2} + {a^2} - {b^2}} \right)}}{{abc}} + \dfrac{{R\left( {{a^2} + {b^2} - {c^2}} \right)}}{{abc}} = \dfrac{{R\left( {{a^2} + {b^2} + {c^2}} \right)}}{{abc}}\left( {dpcm} \right) \)

Bài 16.

Đối với tam giác ABC ta có: \(S = \dfrac{1}{2}ab\sin C = \dfrac{1}{2}{h_C}.c = \dfrac{{abc}}{{4R}} \)

Ta suy ra \({h_c} = \dfrac{{ab}}{{2R}} \). Tương tự ta có \({h_b} = \dfrac{{ac}}{{2R}},{h_a} = \dfrac{{bc}}{{2R}} \)

Do đó:

\(\dfrac{1}{{{h_b}}} + \dfrac{1}{{{h_c}}} = 2R\left( {\dfrac{1}{{ac}} + \dfrac{1}{{ab}}} \right) = 2R\dfrac{{b + c}}{{abc}}\ \)mà $b + c = 2a$

Nên \(\dfrac{1}{{{h_b}}} + \dfrac{1}{{{h_c}}} = \dfrac{{2R.2a}}{{abc}} = \dfrac{{2R.2}}{{bc}} = \dfrac{2}{{{h_a}}} \)

Vậy \(\dfrac{2}{{{h_a}}} = \dfrac{1}{{{h_b}}} + \dfrac{1}{{{h_c}}} \)

Có \(\sin\widehat{A}=\frac{h_c}{b}=\frac{h_b}{c}=\frac{h_c-h_b}{b-c}=\frac{h_b-h_c}{\frac{a}{k}}=\frac{k\left(h_b-h_c\right)}{a}\) (1) 

Lại có : \(\hept{\begin{cases}\sin\widehat{B}=\frac{h_c}{a}\\\sin\widehat{C}=\frac{h_b}{a}\end{cases}}\)\(\Rightarrow\)\(k\left(\sin\widehat{B}-\sin\widehat{C}\right)=\frac{k\left(h_c-h_b\right)}{a}\) (2) 

(1) (2) ... 

\(\sin\widehat{B}=\frac{h_a}{c}\)\(;\)\(\sin\widehat{C}=\frac{h_a}{b}\) (1) 

\(\hept{\begin{cases}\sin\widehat{B}=\frac{h_c}{a}\\\sin\widehat{C}=\frac{h_b}{a}\end{cases}\Leftrightarrow\hept{\begin{cases}h_c=\sin\widehat{B}.a\\h_b=\sin\widehat{C}.a\end{cases}}}\)\(\Rightarrow\)\(k\left(\frac{1}{h_b}-\frac{1}{h_c}\right)=\frac{k}{a}.\left(\frac{1}{\sin\widehat{C}}-\frac{1}{\sin\widehat{B}}\right)\) (2)  

Thay (1) vào (2) ta được \(\frac{k}{a}.\left(\frac{1}{\sin\widehat{C}}-\frac{1}{\sin\widehat{B}}\right)=\frac{k}{a}.\left(\frac{b}{h_a}-\frac{c}{h_a}\right)=\frac{k}{a}.\frac{\frac{a}{k}}{h_a}=\frac{1}{h_a}\)

đpcm