toán hay

B

braga

[TEX]a, \ 5\hat{M}=3\hat{N} \Rightarrow \hat{M}=\frac{3}{5}\hat{N}[/TEX]

Thế vào [TEX]7\hat{M} -4\hat{N} =15^o [/TEX] , ta được:

[TEX]7.\frac{3}{5}\hat{N} -4\hat{N} =15^o \Rightarrow \frac{1}{5}\hat{N}=15^o \Rightarrow \hat{N}=75^o [/TEX]

[TEX]\Rightarrow \hat{M}=\frac{3}{5}.75^o=45^o \Rightarrow \hat{P}=180^o-75-45=60^o[/TEX]

b, N^+P^=180o−80o=100o⇒N^=100o+30o2=65o⇒P^=65o−40o=25ob, \ \hat{N}+\hat{P}=180^o-80^o=100^o \Rightarrow \hat{N}=\frac{100^o+30^o}{2}=65^o \\ \Rightarrow \hat{P}=65^o-40^o=25^o
 
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