Question

Answer 1

\(a,\left\{{}\begin{matrix}a\perp AB\\b\perp AB\end{matrix}\right.\Rightarrow a//b\\ b,a//b\Rightarrow x+y=180^0\left(Trong.cùng.phía\right)\\ \Rightarrow x=180^0-65^0=115^0\\ c,2x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{2}\)

Mà \(x+y=180^0\left(trong.cùng.phía\right)\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x+y}{3+2}=\dfrac{180^0}{5}=36^0\\ \Rightarrow\left\{{}\begin{matrix}x=108^0\\y=72^0\end{matrix}\right.\)