violimpic can gap

C

chonhoi110

Ta có 3 góc $M,N,P$ tỉ lệ thuận với $6,3,4$

\Rightarrow $\dfrac{M}{6}=\dfrac{N}{3}=\dfrac{P}{4}=\dfrac{Q}{a}=\dfrac{M+N+P+Q}{6+3+4+a}=\dfrac{360}{13+a} (1)$

$N-Q=24$

\Rightarrow$ \dfrac{N}{3}=\dfrac{Q}{a}=\dfrac{24}{3-a}(2)$

Từ $(1),(2)$ \Rightarrow$\dfrac{360}{13+a}=\dfrac{24}{3-a}=\dfrac{360+24}{13+a+3-a}=24$

\Rightarrow $M=144^0$
 
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