Tugas / Senin, 5 April 2021
Silahkan di Baca dan di Pahami
'fhalearning.blogspot.com'
Topik : ⇛ Hubungan antara dua sudut
(sub topik : sudut bertolak belakang)
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di tulis di buku catatan MTK
∠ AED = (3x + 20)° = (3 . 15 + 20)° = (45 + 20)° = 65°
∠ BEC = (5x - 10)° = (5 . 15 - 10)° = (75 - 10)° = 65°
(6y + 25)° + 65° = 180°
6y° + 25° + 65° = 180°
6y° + 90° = 180°
6y° = 180° - 90°
6y° = 90°
y° = 90° / 6
y° = 15°
✏
∠ CED = (6y + 25)° = (6. 15 + 25)° = (90 + 25)° = 115°


F. Hubungan Antara Dua Sudut
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Konsep Dasar sudut bertolang belakang -- Penting !!!
∠ 2 = ∠ 4 akan memiliki besar sudut yang sama
dan
∠ 1 + ∠ 2 = 180° (berpelurus)
∠ 3 + ∠ 4 = 180° (berpelurus)
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Diketahui :
∠ AOC = 35°
Ditanya : ∠ AOD, ∠ BOD, dan ∠ BOC
Penyelesaiannya
"analisis gambar diatas"
Jawab :
✏
∠ AOC + ∠ AOD = 180° (berpelurus)
35° + ∠ AOD = 180°
∠ AOD = 180° - 35°
∠ AOD = 145°
✏
∠ BOD = ∠ AOC = 35° (bertolak belakang)
✏
∠ BOC = ∠ AOD = 145° (bertolak belakang)
Jadi; ∠ AOD = 145° , ∠ BOD = 35° , dan ∠ BOC = 145°
jika di gambar seperti berikut;
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Jawab :
✏
3x = 60° (bertolak belakang)
x = 60° / 3
x = 20°
Jadi; nilai x = 20°
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Penyelesaiannya
"analisis gambar diatas"
Jawab :
✏
2x = 120° (bertolak belakang)
x = 120° / 2
x = 60°
✏
4y = 52° (bertolak belakang)
y = 52° / 4
y = 13°
✏
5z + 3 = 68° (bertolak belakang)
5z = 68° - 3
5z = 65°
z = 65° / 5
z = 13°
Jadi; nilai x = 60° , y = 13° , dan z = 13°
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Diketahui :
∠ AOD = (5x + 18)°
∠ AOB = (3x + 18)°
Ditanya : ∠ AOD, ∠ AOB, ∠ BOC dan ∠ COD
Penyelesaiannya
"analisis gambar diatas"
Jawab :
✏
∠ AOC + ∠ AOB = 180° (berpelurus)
(5x + 18)° + (3x + 18)° = 180°
5x° + 18° + 3x° + 18° = 180°
5x° + 3x° + 18° + 18° = 180°
8x° + 36° = 180°
8x° = 180° - 36°
8x° = 144°
x° = 144° / 8
x° = 18°
✏
∠ AOD = (5x + 18)° = (5 . 18 + 18)° = (90 + 18)° = 108°
∠ AOD = (5x + 18)° = (5 . 18 + 18)° = (90 + 18)° = 108°
✏
∠ AOB = (3x + 18)° = (3 . 18 + 18)° = (54 + 18)° = 72°
✏
∠ BOC = ∠ AOD = 108° (bertolak belakang)
✏
∠ COD = ∠ AOB = 72° (bertolak belakang)
Jadi; ∠ AOD = 108° , ∠ AOB = 72° , ∠ BOC = 108° dan ∠ COD = 72°
jika di gambar seperti berikut;
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Diketahui :
∠ AED = (3x + 20)°
∠ CED = (6y + 25)°
∠ BEC = (5x - 10)°
Ditanya : ∠ AED, ∠ CED, ∠ BEC dan ∠ AEB
Penyelesaiannya
"analisis gambar diatas"
Jawab :
✏
∠ AED = ∠ BEC (bertolak belakang)
(3x + 20)° = (5x - 10)°
3x° + 20° = 5x° - 10°
3x° - 5x° = - 20° - 10°
- 2x° = - 30°
x° = - 30° / -2
x° = 15°
✏∠ AED = (3x + 20)° = (3 . 15 + 20)° = (45 + 20)° = 65°
∠ BEC = (5x - 10)° = (5 . 15 - 10)° = (75 - 10)° = 65°
✏
∠ CED + ∠ BEC = 180° (berpelurus)(6y + 25)° + 65° = 180°
6y° + 25° + 65° = 180°
6y° + 90° = 180°
6y° = 180° - 90°
6y° = 90°
y° = 90° / 6
y° = 15°
✏
∠ CED = (6y + 25)° = (6. 15 + 25)° = (90 + 25)° = 115°
✏
∠ AEB = ∠ CED = 115° (bertolak belakang)
∠ AEB = ∠ CED = 115° (bertolak belakang)
Jadi; x° = 15°, y° = 15°, ∠ AED = 65° , ∠ BEC = 65° , ∠ CED = 115° dan ∠ AEB = 115°
jika di gambar seperti berikut;
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