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Question
secondary 4 | A Maths
3 Answers Below
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Pls help in this qn thank you :)
Date Posted: 4 years ago
Views: 255
Will have a look at this R-formula question by 4 am.
OAB = θ
AB = 17 cos θ
OB = 17 sin θ
OC = 31 cos θ
CD = 31 sin θ
BC = OC - OB
= 31 cos θ - 17 sin θ
AB + BC + CD
= 17 cos θ + 31 cos θ - 17 sin θ + 31 sin θ
= ( 48 cos θ + 14 sin θ ) cm
here onwards,
u use R formula :
48 cos θ + 14 sin θ = R cos( θ - α )
for part (ii)
(iii)
max value is R
.
.
.
AB = 17 cos θ
OB = 17 sin θ
OC = 31 cos θ
CD = 31 sin θ
BC = OC - OB
= 31 cos θ - 17 sin θ
AB + BC + CD
= 17 cos θ + 31 cos θ - 17 sin θ + 31 sin θ
= ( 48 cos θ + 14 sin θ ) cm
here onwards,
u use R formula :
48 cos θ + 14 sin θ = R cos( θ - α )
for part (ii)
(iii)
max value is R
.
.
.
See 3 Answers
done
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Good evening Carolyn! Here are my workings for the proving part.
Date Posted:
4 years ago
done
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Good evening Carolyn! Here are my workings for the second part.
Date Posted:
4 years ago
done
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Good evening Carolyn! Here are my workings for the third part.
Date Posted:
4 years ago
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