Question

Hitung sudut terkecil dari jarum jam berikut
a. pukul 04.30
b.pukul 07.20
c.pukul 05.12
d.pukul19.01
e pukul 10.40
tolong jawab

Answer 1

A. 04.30
=(6 × 30) - (4 × 30 + 30/60 × 30)
=180 - (120 + 15)
=180 - 135
=45
b. 07.20
=(7 × 30 + 20/60 × 30) - (4 × 30)
=(210 + 10) - 120
=220 - 120
=100
c. 05.12
=(5 × 30 + 12/60 × 30) - (2,4 × 30)
=(150 + 6) - 72
=156 - 72
=84
d.19.01
=(7 × 30 + 1/60 × 30) - (0,2 × 30)
=(210 + 0,5) - 6
=210,5 - 6
=204,5
e. 10.40
=(10 × 30 + 40/60 × 30) - (8 × 30)
=(300 + 20) - 240
=320-240
=80