Matematika

Pertanyaan

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1 Jawaban

  • ∠DEC+∠ECD+∠CDE=180°
    110°+30°+∠CDE=180°
    ∠CDE=180°-140°
    ∠CDE=40°

    ∠BEP+∠DEC=180°
    ∠BEP=180°-110°
    ∠BEP=70°

    ∠BPC=2∠CDE
    ∠BPC=2(40°)
    ∠BPC=80°

    ∠BPE+∠BPC=180°
    ∠BPE=180°-80°
    ∠BPE=100°

    ∠EBP+∠BPE+∠BEP=180°
    ∠EBP=180°-100°-70°
    ∠EBP=10°

    ∠BAC=∠CDE=40°

    ∠AEB=∠DEC=110°

    ∠AEB+∠BAC+∠ABE=180°
    ∠ABE=180°-110°-40°
    ∠ABE=30°

    ∠ABP=∠ABE+∠EBP
    ∠ABP=30°+10°
    ∠ABP=40°