Bantu jawab pakek caranya kak! jan ngasal:v
byburbtehethqryhrynryqnrqyjqrynqrggn
Jawab:
1. ∠SQT + ∠TQR = 90°
x + 28° + 6x – 15° = 90°
7x + 13° = 90°
7x = 90° – 13°
7x = 77°
x = 77°/7
x = 11°
∠SQT = x + 28° = 11° + 28° = 39°
∠TQR = 6x – 15° = 6(11°) – 15° = 66° – 15° = 51°
sudut yang saling berpenyiku adalah ∠SQT dan ∠TQR.
2. a. 5b + 10° + 80° + 15b – 20° = 180°
20b + 70° = 180°
20b = 180° – 70°
20b = 110°
b = 110°/20°
b = 5,5°
b. 46° + b + 29° + 5b + 15° = 180°
6b + 90° = 180°
6b = 180° – 90°
6b = 90°
b = 90°/6
b = 15°
3. a. misal suatu sudut = A, penyikunya = B
A = 4B
A + B = 90°
4B + B = 90°
5B = 90°
B = 90°/5 = 18°
sudut tersebut A = 4B = 4(18°) = 72°
b. pelurus = 132°
misal sudut tersebut A, maka
A + 132° = 180°
A = 180° – 132°
A = 48°
penyiku sudut A = 90° – 48° = 42°
4. 2x = 120° (saling bertolak belakang)
x = 120°/2
x = 60°
3y = 42° (saling bertolak belakang)
y = 42°/3
y = 14°
5z + 3 = 68° (saling bertolak belakang)
5z = 68° – 3
5z = 65°
z = 65°/5
z = 13°
5. ∠DEA = ∠CEF (saling bertolak belakang)
∠DEA = 102°
∠DEA + ∠BAE = 180° (sudut dalam sepihak)
102° + 3y = 180°
3y = 180° – 102°
3y = 78°
y = 78°/3
y = 26°