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Posted by says it all (U15256656) on Sunday, 6th May 2012
Can someone please help to do this question from the Official SQA Past Papers Mathematics Intermediate 2 question 12 paper 2 question a and b?
Posted by Maths Teacher BBC (U15178780) on Tuesday, 8th May 2012
The arms on a wind turbine rotate at a steady rate.....
(a) At a time of 30 seconds then t must = 30
H = 8 + 4 sin t
H = 8 + 4 sin 30
H= 10 metres
(b) At a height of 10.5 m then h = 10.5
10.5 = 8 + 4 sin t
8 + 4 sin t = 10.5
4 sin t = 10.5 – 8
4 sin t = 2.5
Sin t = 2.5/4
Sin t = 0.625
T = inverse sin 0.625
T = 38.7 seconds
As you want two times you must put this into a CAST diagram:
So you want sin to be positive so it lies in A and S
In A = 0 + 38.7 = 38.7 seconds
In S = 180 – 38.7 = 141.3 seconds
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