Question

Cho tam giác ABC có b-c=\(\dfrac{a}{2}\)

a, SinA=2sinB-2sinC

b, \(\dfrac{1}{h_a}=\dfrac{1}{h_b}-\dfrac{1}{h_c}\)

Answer 1

\(a=2b-2c\Rightarrow sinA.2R=2sinB.2R-2sinC.2R\)

\(\Rightarrow sinA=2sinB-2sinC\)

\(ah_a=bh_b=ch_c\Rightarrow\left(2b-2c\right)h_a=bh_b=ch_c\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{h_a}=\dfrac{2b-2c}{b}.\dfrac{1}{h_b}\\\dfrac{1}{h_a}=\dfrac{2b-2c}{c}.\dfrac{1}{h_c}\end{matrix}\right.\) 

\(\Rightarrow\dfrac{1}{h_a}=\dfrac{1}{h_b}-\dfrac{1}{h_c}+\left(\dfrac{b}{c.h_c}-\dfrac{c}{b.h_b}\right)\)

Câu này đề sai tiếp, biểu thức \(\dfrac{b}{c.h_c}-\dfrac{c}{b.h_b}\) kia không thể bằng 0