∵∠P=180°-∠PBE-∠PEB, ∠PBE=(180°-∠C-∠CGB)①, ∠PEB=180°-∠D-∠DAE②, ∠EAG=(180°-∠DAG)③ ∴∠P=180°-∠PBE-∠PEB =180°-(180°-∠C-∠CGB)-∠PEB =90°+∠C+∠CGB-(180°-∠D-∠DAE) =∠C+∠CGB-90°+∠D+(∠DAG+∠CAM) =∠C+∠CGB-90°+∠D+(180°-∠D-∠DGA)+(∠D+∠DGA) =90°+∠C+∠D =90°+×62°+×30° =136°. 故答案为:136°.
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