Eric Nicholas K's answer to John tay's Secondary 4 A Maths Singapore question.
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First two parts
Date Posted:
4 years ago
Lol this question again
Alternative for ii)
When t = 3, the hour hand is horizontal and pointing rightwards.
The hand at this position makes an angle of π/2 rad with its previous position at the 12am mark.
Then h = 25
25 = 5(5 - 2 cos kt) = 25 - 10 cos kt
10 cos kt = 0 → coskt = 0
kt = π/2 rad
(We do not pick other bigger angles as the angle change from the 12am mark is not obtuse/is the magnitude of a right angle)
Since t = 3hr
3k hr = π/2 rad
k = π/6 rad/h
When t = 3, the hour hand is horizontal and pointing rightwards.
The hand at this position makes an angle of π/2 rad with its previous position at the 12am mark.
Then h = 25
25 = 5(5 - 2 cos kt) = 25 - 10 cos kt
10 cos kt = 0 → coskt = 0
kt = π/2 rad
(We do not pick other bigger angles as the angle change from the 12am mark is not obtuse/is the magnitude of a right angle)
Since t = 3hr
3k hr = π/2 rad
k = π/6 rad/h
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