# Peee twoo question...

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Thread starter 15 years ago
#1
Hey.. I'm a bit stuck on this one, don't know what to do:
In the green heinemann book, review ex 2, p.176, question 47

There's a diagram which you might need to see to answer the question, but here's the question anyway

A semicircle, has midpoint O of the diameter AB.
The point P is on the semicircle such that the area of sector POB is equal to the area enclosed by the segment AP

The angle POB is x radians
Prove that x = (1/2)(pi - sin x)

anyone?
0
15 years ago
#2
Area of shaded segment = area of sector AOP - area of triangle AOP

= 1/2(pi - x)r^2 - 1/2(r^2)(sin (pi - x))
= r^2/2 (pi-x - sin (pi-x))

Area of sector POB = 1/2(x)(r^2)

r^2/2 (pi-x - sin (pi-x)) = 1/2(x)(r^2)
pi - x - sin (pi - x) = x
2x = pi - (sin (pi-x))
x = 1/2 (pi - sin x)

I think?
0
15 years ago
#3
(Original post by mik1a)
Area of shaded segment = area of sector AOP - area of triangle AOP

= 1/2(pi - x)r^2 - 1/2(r^2)(sin (pi - x))
= r^2/2 (pi-x - sin (pi-x))

Area of sector POB = 1/2(x)(r^2)

r^2/2 (pi-x - sin (pi-x)) = 1/2(x)(r^2)
pi - x - sin (pi - x) = x
2x = pi - (sin (pi-x))
x = 1/2 (pi - sin x)

I think?
guys we need to know p1 formulas too? i have never seen a paper where we really need p1 formulas
0
15 years ago
#4
(Original post by TheWolf)
guys we need to know p1 formulas too? i have never seen a paper where we really need p1 formulas
I couldn't think of any other way.
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Thread starter 15 years ago
#5
Aah, of course 1/2 ab sin C

Cheers!
0
Thread starter 15 years ago
#6
(Original post by TheWolf)
guys we need to know p1 formulas too? i have never seen a paper where we really need p1 formulas
I've been told that you're expected to have P1 knowledge...
0
15 years ago
#7
(Original post by kimoni)
I've been told that you're expected to have P1 knowledge...
CRAP i cant remember most of the p1 formulas :P
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Thread starter 15 years ago
#8
(Original post by TheWolf)
CRAP i cant remember most of the p1 formulas :P
me neither.. Although I've just learnt the circle ones from doing this question.
I sincerely doubt we'll be asked any sequences and series questions, which contained those AP and GP forumlae...
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Thread starter 15 years ago
#9
(Original post by lgs98jonee)
wot have u done here??have u said that sin(pi-x)=sin x that isnt true is it? or am i missing sumthing simple ??? mik1a have u even done p2 yet? - as in started it at school??
If you havent done P2 you won't get this...
sin(pi - x) expands to sin(pi)cos(x) - cos(pi)sin(x)
and sin(pi) is 0....
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Thread starter 15 years ago
#10
(Original post by lgs98jonee)
erm no actually sinpi=1, so that goes to cosx as cospi=0
sin(pi) is 0, not 1... and cos(pi) is -1, read it through, and it'll make sense!
0
15 years ago
#11
(Original post by kimoni)
sin(pi) is 0, not 1... and cos(pi) is -1, read it through, and it'll make sense!
ah yes good point..keep think pi radians=90 degrees
0
Thread starter 15 years ago
#12
(Original post by lgs98jonee)
ah yes good point..keep think pi radians=90 degrees
heh, just don't make that mistake in the exam!
0
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