IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: halosh92 on March 11, 2010, 06:46:41 pm
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1)triangle ABC is such that AB=5cm , AC=10 cm and triangle ABC=90 degrees. (angle B=90)
an arc of a circle , center A and radius 5cm cuts AC at D
a) state , in radians the value of angle A in BAC
b) calculate the are of the region enclosed by BC , DC , and the arc BD
2) a minor sector OMN of a circle center O and radius "r". the perimeter of the sector is 100cm and the area of the sector is A cm^2
a) show that A=50r-r^2
b) given that r varies , find:
i) the value of "r" for which A is a maximum and show that A is a maximum
ii) the value of MON for this maximum area
iii) the maximum area of sector OMN
plzzzz help!
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2 hours
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3) sector OAB of a circle, center O and radisu "r" cm. the length of the arc AB is "p" cm and AOB is "tita" radians..
a) find "tita" in terms of "p" and 'r"
b) deduce that the area of the sector is 1/2pr
given that r= 4.7 and p=5.3, where each has been measured to 1 decimal place, find giving your answer to 3 decimal places.
c) the least possible value of the area of the sector
d) the range of possible values of "tita"
i dono how to do part c and d
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1)triangle ABC is such that AB=5cm , AC=10 cm and triangle ABC=90 degrees. (angle B=90)
an arc of a circle , center A and radius 5cm cuts AC at D
a) state , in radians the value of angle A in BAC
b) calculate the are of the region enclosed by BC , DC , and the arc BD
a)pi/3
b)1/2(5*sqrt(75)-5^2)
2) a minor sector OMN of a circle center O and radius "r". the perimeter of the sector is 100cm and the area of the sector is A cm^2
a) show that A=50r-r^2
b) given that r varies , find:
i) the value of "r" for which A is a maximum and show that A is a maximum
ii) the value of MON for this maximum area
iii) the maximum area of sector OMN
a)100=2r+r theta
theta=(100-2r)/r
A=1/2r^2 theta=1/2*r^2*(100-2r)/r=50r-r^2
b)dA/dr=50-2r=0 so r=25
d^2A/dr" =-2<0 hence max
theta=MON=(100-2r)/r=2 when r=25
Area=50r-r^2 =625 when r=25
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c)find 1/max(2pr)=1/(2*4.65*6.35)
d)tita=r/r
find (max p)/(min r) to (min p)/(max r)
ie 5.35/4.65 to 5.25/4.75
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1)triangle ABC is such that AB=5cm , AC=10 cm and triangle ABC=90 degrees. (angle B=90)
an arc of a circle , center A and radius 5cm cuts AC at D
a) state , in radians the value of angle A in BAC
b) calculate the are of the region enclosed by BC , DC , and the arc BD
a)pi/3
b)1/2(5*sqrt(75)-5^2)
2) a minor sector OMN of a circle center O and radius "r". the perimeter of the sector is 100cm and the area of the sector is A cm^2
a) show that A=50r-r^2
b) given that r varies , find:
i) the value of "r" for which A is a maximum and show that A is a maximum
ii) the value of MON for this maximum area
iii) the maximum area of sector OMN
a)100=2r+r theta
theta=(100-2r)/r
A=1/2r^2 theta=1/2*r^2*(100-2r)/r=50r-r^2
b)dA/dr=50-2r=0 so r=25
d^2A/dr" =-2<0 hence max
theta=MON=(100-2r)/r=2 when r=25
Area=50r-r^2 =625 when r=25
for question one, i didnt understand how you got the area of the region enclosed by BC , DC , and the arc BD
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When I get up I will draw a diagram
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here
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THX ALOT! ;D
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Solve, for 0 ? x < 360, the equation
3 cos2 x° + sin2 x° + 5 sin x° = 0.
how do u come about such questions???????/ step by step explanation plzzzz
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2 hours
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a circle of radius 12 cm which passes through the points P and Q.
The chord PQ subtends an angle of 120° at the centre of the circle.
(a) Find the exact length of the major arc PQ.
(b) Show that the perimeter of the shaded minor segment is given by
k(2pie + 3sqrt 3 ) cm, where k is an integer to be found.
(c) Find, to 1 decimal place, the area of the shaded minor segment as a percentage
of the area of the circle.
i just cant get part (B)
how do u find it for the chord? any specific formula isnt it that we have to us ethe cosine rule???????
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olve, for 0 ? x < 360, the equation
3 cos2 x° + sin2 x° + 5 sin x° = 0.
are you sure this is the question?
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Perimeter =length of chord =sqrt(2r^2(1-cos theta))+r theta =r sqrt(2(1-cos 120))+r*2pi/3=12(sqrt3 +2pi/3)=4(sqrt3+2pi)
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olve, for 0 ? x < 360, the equation
3 cos2 x° + sin2 x° + 5 sin x° = 0.
are you sure this is the question?
0 < x < 360, the equation
3 cos2 x° + sin2 x° + 5 sin x° = 0.
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are they cos^2 x sin^2 x etc
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solve the following equations for thita, in the interval 0<thita<360
a) sin thita + cos thita = 0
b) (sin thita - 1)(5 cos thita + 3) = 0
solve the following equations for "x" , giving your answers to 3sf where appropriate, in the intervals indicated:
a) tan x = - sqrt3 /3 0<x<720
b) tan x = 2.90 80<x<440
solve in the intervals indicated . the following equations for thita, where thita is measured in radians,
give your answer in terms of pie or 2pie or 2 decimal places.
a) sin thita =0 -2pie<thita<2pie
b) 2 cos thita = 3 sin thita 0<thita<2pie
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yep
sorry typing error
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a)sin x +cos x=0
sin x =-cos x
tan x =-1
x=-45,180-45, 360-45 etc
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(sin x-)(5cos x+3)=0
sinx-1=0 so sinx =1 so x=90, 180-90, 360+90, etc
only 90 in range
5cos x+3=0
cos x=-3/5 so x=126.9
x=126.9,360-126.9, 360+126.9, etc
only 1st two in range
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for the a and b tan questions find any answer to tanx=c then keep adding 180
sinx=0 x=-360,-180,0,180,360
2cos x=3 sinx so tanx=2/3 so x=33.7,180+33.7,360+33.7 etc
Can you post the very first question with all the squares made clear
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for the a and b tan questions find any answer to tanx=c then keep adding 180
sinx=0 x=-360,-180,0,180,360
2cos x=3 sinx so tanx=2/3 so x=33.7,180+33.7,360+33.7 etc
Can you post the very first question with all the squares made clear
for any tan questions !!-do we ALWAYS add 180???
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yes always add 180
http://www.astarmathsandphysics.com/a_level_maths_notes/C3/a_level_maths_notes_c3_finding_multiple_solutions_of_trigonometric_equations.html
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okay thx alot ;D ;D ;D ;D
binomial expansion :
(when (1-3/2 x)^p is expanded in ascending powers of "x" , the coefficient of "x" is -24
a) find the value of "p"
b) find the coefficient of x^2 in the expansion
c) find the coefficient of x^3 in the expansion
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(1-3x/2)^p =1+(-3px/2)+(p(p-1)/2!*(-3x/2)^2+(p(p-1)(p-2)/3!)(-3x/2)^3+....
-3p/2=-24 so p=16
coefficienct of x^2 (p(p-1)/2!*(-3/2)^2 =(16(16-1)/2!*(-3/2)^2 =270
coefficient of x^3 (p(p-1)(p-2)/3!)(-3/2)^3 =(16(16-1)(16-2)/3!)(-3/2)^3 =1260
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the vertices of a quadrilateral ABCD has coordinates A(-1,5) B(7,1) C(5,-3) D(-3,1)
show that the quadrileteral is a triangle
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You mean a parallelogram. In morning
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You mean a parallelogram. In morning
sorry i meant rectangle not paralleolgram
:P
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A(-1,5) B(7,1) C(5,-3) D(-3,1)
AB=(7,1)-(-1,5)=(8,-4)
C-D=(5,-3)-(-3,1)=(8,-4)=B-A so Ab is same length as CD
CB=(7,1)-(5,-3)=(2,4)=DA
Also AB.CB=8*2+-4*4=0 etc
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find the coordinates of the point where the tangent to the curve y=x^2 + 1 at the point (2,5)
meets the normal to the same curve at the point (1,2)
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1 hour
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find the coordinates of the point where the tangent to the curve y=x^2 + 1 at the point (2,5)
meets the normal to the same curve at the point (1,2)
tangent to curve y=x^2 + 1 dy/dx=2x=2*2=4
y-5=4(x-2) so y=4x-3 (1)
normal at (1,2) has gradient -1/2*1=-1/2
y-2=(-1/2)(x-1) so y=-0.5x+2.5 (2)
(1) -(2) gives 0=4.5x-5.5 so x=9/11 and from (1) y=4*9/11 -3=3/11
(9/11,3/11)
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find the coordinates of the point where the tangent to the curve y=x^2 + 1 at the point (2,5)
meets the normal to the same curve at the point (1,2)
tangent to curve y=x^2 + 1 dy/dx=2x=2*2=4
y-5=4(x-2) so y=4x-3 (1)
normal at (1,2) has gradient -1/2*1=-1/2
y-2=(-1/2)(x-1) so y=-0.5x+2.5 (2)
(1) -(2) gives 0=4.5x-5.5 so x=9/11 and from (1) y=4*9/11 -3=3/11
(9/11,3/11)
how did you find the gradient of the normal..on wat basis did u come up with these equation? (normal at (1,2) has gradient -1/2*1=-1/2
y-2=(-1/2)(x-1) so y=-0.5x+2.5 )
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Dy/dx=2x and x=1 so dy/dx=2 and gradient of normal=-1/2