Mathematics: Post your doubts here!

[Translucent231]

View attachment 21800

After you've drawn the argand diagram it's just a bit of geometry. The red line is the Max Re z which is 2+x. You have to find the value of x. Consider the right-angled triangle with the hypotenuse 2 as it is the radius of the circle. The angle subtended at the centre is π/4, so now just find the length of opposite. The opposite is x so Re z is 2+opp

lol.. why have you ignored the yellow area on the right side of the red line?

 

[littlecloud11]

lol.. why have you ignored the yellow area on the right side of the red line?

Ignored? Didn't get your question. The entire yellow region is the shaded area, the red line merely indicates the greatest real value in the possible region.

 

[imalikshake]

View attachment 21800

After you've drawn the argand diagram it's just a bit of geometry. The red line is the Max Re z which is 2+x. You have to find the value of x. Consider the right-angled triangle with the hypotenuse 2 as it is the radius of the circle. The angle subtended at the centre is π/4, so now just find the length of opposite. The opposite is x so Re z is 2+opp

Oh! Man, that was simply Thank you so much for the help!

 

[littlecloud11]

Oh! Man, that was simply Thank you so much for the help!

No Problem!

 

[Translucent231]

Ignored? Didn't get your question. The entire yellow region is the shaded area, the red line merely indicates the greatest real value in the possible region.

yaar i dont understand how u can say that the red line shows the greatest real value in the possible region.... the real value is the X coordinate on the argand diagram yes? ... and clearly there are greater X coordinates than the red line...... in this case the greatest real value of Z should be 4 ... not 2 + rt2 ......you must have misread the question because the mistake you made is that you shaded the region for which the arg(z) > -pi/4 .... u had to shade it for arg(z) < -pi/4 ....... -2 is LESS than -1 .. -1 is GREATER than -3 ....

 

[Tkp]

hy can any1 help me in m1 oct nov 09-3,4,5.
pls pls pls.

 

[PhyZac]

hy can any1 help me in m1 oct nov 09-3,4,5.
pls pls pls.

Assalamu Alikum
which variant?

 

[Tkp]

Assalamu Alikum
which variant?

its variant 1

 

[PhyZac]

its variant 1

Okay, i will take time for that , later i will post if possible In Shaa Allah

 

[PhyZac]

its variant 1

Question 3

Resolve (>) and (^)
for (>) Q-Pcos(60) = 12cos80 (equation 1 )
for (^) Psin60 = 12sin80 (equation 2)
find P from equation 1 ... substitute to equation 1 and find Q

 

[Tkp]

Question 3

Resolve (>) and (^)
for (>) Q-Pcos(60) = 12cos80 (equation 1 )
for (^) Psin60 = 12sin80 (equation 2)
find P from equation 1 ... substitute to equation 1 and find Q

how come is 12 sin 80.cn u explain it

 

[tanmaydube]

Help please!
Summer 2008 Question 3 4 6 8
little help would also count...

Thank You

please help ... any question any part would do please...

 

[littlecloud11]

please help ... any question any part would do please...

3i) Perimeter of rectangle = a+3a+a+3a = 8a
Perimeter of Sector= r+r+rx = 2r+rx
Now express r in terms of a. Consider triangle DAN, the angle DAN is (π/2-x)
Cox (π/2-x) = a/r
[Cox (π/2-x) can be written as sinx]
so, sinx =a/r, r= a/sinx
Perimeter of sector = 1/2 perimeter of rectangle
2r+ rx = 1/2* 8a
substitute r= a/sinx in the equation and cancel 'a' from both sides. you get-
2/sinx + x/sinx = 4
or, (2+x)/sinx =4 so sinx= (2+x)/4

ii) Substitute the X1 value in the given equation in place of Xn to obtain X2. Then use the X2 value in place of Xn to obtain X3... and so on.
X1=.8
X2= sin^-1 (2+.8/4) = .7754
X3= sin^-1 (2+.7754/4) = .7668
X4= sin^-1 (2+.7668/4) = .7638
X3 and X4 coincide till 2 decimal places. So your answer is .76

 

[PhyZac]

how come is 12 sin 80.cn u explain it

Asslamu Alikum..

See you are resolving upward...and you know the final vector...the final or resultant is the 12 vector...!
So if you check the only vector tht forms an upward is P vector in the form of Psin60 and to resolve the upward of 12 vector u take the 12sin80...i cant explain why we took 12 sin 80...but check this video..and u myt get the point
http://www.examsolutions.net/maths-revision/mechanics/forces/resultant/three-forces/tutorial-1.php

 

[PhyZac]

Asslamu Alikm waRahmatullahi Wa Barakatoho littlecloud11 Dug
...Can anyone show how to solve this question...?

The answer is
106.3 is the angle
and component bit is 7.2N

 

[Tkp]

Asslamu Alikm waRahmatullahi Wa Barakatoho
...Can anyone show how to solve this question...?

The answer is
106.3 is the angle
and component bit is 7.2N

R^2=P^2+Q^2+2PQCOS theta
u will get the angle
i have prblm in the 2nd 1.like the ans would be 10+10 costheta.but y 10 iis added here.they told to find the vector in the direction of OA

 

[PhyZac]

R^2=P^2+Q^2+2PQCOS theta
u will get the angle
i have prblm in the 2nd 1.like the ans would be 10+10 costheta.but y 10 iis added here.they told to find the vector in the direction of OA

What is R? P? Q?

Second is fine....
u find R(>) tht is 10 - 10 cos (180-theta) or 12 cos (theta/2)

 

[littlecloud11]

Asslamu Alikm waRahmatullahi Wa Barakatoho littlecloud11 Dug
...Can anyone show how to solve this question...?

The answer is
106.3 is the angle
and component bit is 7.2N

There might be an easier way to solve this but this is how i did it.


Split the Left hand 10N force into its vertical and horizontal components by breaking θ into 90 degree and (θ-90).
Then resolve the vertical and horizontal component.
There is only one vertical component 10 Cos (θ-90) which is to 10Sinθ
The horizontal component would be 10- 10 sin(θ-90) or 10-10 Cosθ
Now consider these the vertices of a right angled triangle with the hypotenuse as 12N.
So, 12^2 =(10 Sinθ)^2 + (10-10Cosθ)^2
144 = 100 sin^2θ + 100 - 200 Cosθ +100 Cos^2θ
144 =100 - 200 Cosθ + 100 (sin^2θ + Cos^2θ)
144 = 100 - 200 Cosθ + 100
Solve and you get θ as 106.3

 

[Tkp]

What is R? P? Q?

Second is fine....
u find R(>) tht is 10 - 10 cos (180-theta) or 12 cos (theta/2)

.wtf.why can i upload the file
well r is the resultant one.take p as x and q as y .then solve it

 

[PhyZac]

There might be an easier way to solve this but this is how i did it.
View attachment 21926

Split the Left hand 10N force into its vertical and horizontal components by breaking θ into 90 degree and (θ-90).
Then resolve the vertical and horizontal component.
There is only one vertical component 10 Cos (θ-90) which is to 10Sinθ
The horizontal component would be 10- 10 sin(θ-90) or 10-10 Cosθ
Now consider these the vertices of a right angled triangle with the hypotenuse as 12N.
So, 12^2 =(10 Sinθ)^2 + (10-10Cosθ)^2
144 = 100 sin^2θ + 100 - 200 Cosθ +100 Cos^2θ
144 =100 - 200 Cosθ + 100 (sin^2θ + Cos^2θ)
144 = 100 - 200 Cosθ + 100
Solve and you get θ as 106.3

Jazaki Allah khairan!! Thank you so much for your help! , Alhamdulilah now i get it! May Allah reward you. May Allah provide you with highest grades, in this life and hereafter...!! I think your way is easier than markscheme way!