Question

Tam giác ABC có Góc A = 2 lần góc B , 5 lầm góc B = 4 lần góc C . Kẻ CH vuông góc AB ( H E AB) . Góc ACH = ?   mik đang cần gấp , làm theo l7 nhé cảm ơn nhiều

 

Answer 1

\(5\cdot\widehat{B}=4\cdot\widehat{C}\)

=>\(\widehat{C}=\dfrac{5}{4}\cdot\widehat{B}\)

Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)

=>\(2\cdot\widehat{B}+\widehat{B}+\dfrac{5}{4}\cdot\widehat{B}=180^0\)

=>\(\widehat{B}=180^0:\dfrac{17}{4}=\dfrac{720^0}{17}\)

\(\widehat{C}=\dfrac{5}{4}\cdot\dfrac{720^0}{17}=\left(\dfrac{900}{17}\right)^0\)

\(\widehat{A}=2\cdot\widehat{B}=2\cdot\dfrac{720}{17}=\dfrac{1440}{17}^0\)
ΔACH vuông tại H

=>\(\widehat{HAC}+\widehat{HCA}=90^0\)

=>\(\widehat{HCA}=90^0-\left(\dfrac{1440}{17}\right)^0=\left(\dfrac{90}{17}\right)^0\)