`\hat{N}:\hat{P}=3:5`
`<=>\hat{N}/\hat{P}=3/5`
`<=>\hat{N}=3/5\hat{P}`
Vì tổng `3` góc trong một tam giác bằng `180^o` nên:
`\hat{NMP}+\hat{N}+\hat{P}=180^o`
`<=>100^o +3/5\hat{P}+\hat{P}=180^o`
`<=>100^o +8/5\hat{P}=180^o`
`<=>8/5\hat{P}=180^o-100^o`
`<=>8/5\hat{P}=80^o`
`<=>\hat{P}=80^o:8/5`
`<=>\hat{P}=50^o`
`=>\hat{N}=3/(5).50^o=30^o`
Vậy `\hat{N}=30^o, \hat{P}=50^o`