# Mathematics: Post your doubts here!

#### Attachments

• 9709_w13_qp_32.pdf
90.1 KB · Views: 9

#### Attachments

• Untitled.png
154.7 KB · Views: 8

HELP!! IN PART B

#### qwertypoiu

Theta = €
angle BOC = 2π-4€
AB = sqrt(r^2 + r^2 - 2(r)(r)cos(π-2€) )
= sqrt(2r^2(1-cos(π-2€)))
SECTOR OBAC = 1/2*r^2 * (2π-4€) = r²(π-2€)
Sector ABC = 1/2*(2r^2)(1-cos(π-2€))*(2€) = 2€r²(1-cos(π-2€))

Shaded area = r²(π-2€) +2€r²(1-cos(π-2€)) - r²(sin(π-2€))
=πr² - 2£r²cos(π-2€) - r²sin(π-2£)

πr² - 2£r²cos(π-2€) - r²sin(π-2£) = 1/2*π*r²
2£r²cos(π-2€) + r²sin(π-2£) = ½πr²
2£cos(π-2€) + sin(π-2£) = ½π
2€(-cos(2€)) + sin(2€) = ½π
-4€cos(2€) + 2sin(2€) = π
4€cos(2€) - 2sin(2€) = -π
4€cos(2€) = 2sin(2€) - π
cos(2€) = [2sin(2€) - π ] / 4€

#### Serial_Ripper

Theta = €
angle BOC = 2π-4€
AB = sqrt(r^2 + r^2 - 2(r)(r)cos(π-2€) )
= sqrt(2r^2(1-cos(π-2€)))
SECTOR OBAC = 1/2*r^2 * (2π-4€) = r²(π-2€)
Sector ABC = 1/2*(2r^2)(1-cos(π-2€))*(2€) = 2€r²(1-cos(π-2€))

Shaded area = r²(π-2€) +2€r²(1-cos(π-2€)) - r²(sin(π-2€))
=πr² - 2£r²cos(π-2€) - r²sin(π-2£)

πr² - 2£r²cos(π-2€) - r²sin(π-2£) = 1/2*π*r²
2£r²cos(π-2€) + r²sin(π-2£) = ½πr²
2£cos(π-2€) + sin(π-2£) = ½π
2€(-cos(2€)) + sin(2€) = ½π
-4€cos(2€) + 2sin(2€) = π
4€cos(2€) - 2sin(2€) = -π
4€cos(2€) = 2sin(2€) - π
cos(2€) = [2sin(2€) - π ] / 4€
THANKS!

#### manya

someone tell me how to write equation of plane in the form 3x+2y-3z+1=0 to the form r.n=a.n

#### eliyeap

someone tell me how to write equation of plane in the form 3x+2y-3z+1=0 to the form r.n=a.n
Take r as (x,y,z) and multiply with your normal

#### Georgtchernev45

can anyone do questions 5 part ii) and 6 ii) from the Oct/Nov 2014 paper 32

Need Help..?

#### musiclover gurl

Once again, sorry for the late reply, was really busy with revision. Uhmm, the first two part of the question is a bit blurry. Could u just type it again...? Please?
Ya, AS level

#### The Sarcastic Retard

Once again, sorry for the late reply, was really busy with revision. Uhmm, the first two part of the question is a bit blurry. Could u just type it again...? Please?
Ya, AS level
(i) OB = 6 cm
AO = 6 cm
AB = sqrt(36 + 36)
AB = 6sqrt(2)

(ii) tan^-1 (6/6) = pi/4

#### musiclover gurl

(i) OB = 6 cm
AO = 6 cm
AB = sqrt(36 + 36)
AB = 6sqrt(2)

(ii) tan^-1 (6/6) = pi/4
Thanks a lot!

#### Wolfgangs

A ball is kicked with an initial speed v0=10 m/s at an angle of θ= 37◦above the horizontal. Its speed in m/s at maximum height is:

#### Wolfgangs

Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?

#### Wolfgangs

Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?

#### Farheen1201

A ball is kicked with an initial speed v0=10 m/s at an angle of θ= 37◦above the horizontal. Its speed in m/s at maximum height is:
10cos(37) = 7.98 m/s

Ali pushes a block by applying a force of F = -10i + 10j in units of newtons. What is the direction of this force (the angle of the vector with the +x direction)?
I am guessing its 45

#### nehaoscar

Any revision notes for P3 and S1?