maths c2 edexcel

[halosh92]

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!

[astarmathsandphysics]

2 hours

[halosh92]

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

[astarmathsandphysics]

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
 

[astarmathsandphysics]

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

[halosh92]

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

[astarmathsandphysics]

When I get up I will draw a diagram

[astarmathsandphysics]

here

[halosh92]

THX ALOT!

[halosh92]

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

[astarmathsandphysics]

2 hours

[halosh92]

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?

[astarmathsandphysics]

olve, for 0 ? x < 360, the equation
3 cos2 x° + sin2 x° + 5 sin x° = 0.

are you sure this is the question?

[astarmathsandphysics]

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)

[halosh92]

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.