Mathematics: Post your doubts here!

[Dug]

Thank you very much! And unfortunately no one solved the other one =P

I can't solve the whole question but I am sure this will help.

y=x/2 + 2
For the coordinates of A, put x=0 in the equation and you get y = 2 therefore A(0,2)
The gradient of BD = -2
D(10,-3)
-3=10(-2) + c
c=17
We got the equation of BD ;- y=-2x + 17

For the coordinates of B, set the 2 equations equal and solve.

As the question says AB=BC, it means B is the mid-point of the line.
We have the coordinates of A and B right?
xB , yB = 0 + xC /2 , 2 + yC/2
xC/2 = xB and 2+yC/2 = yB
For this part, you could have used the vector move too.

I may have missed something because I am in a hurry but the method is the same.

 

[2pac]

I'll do the ii(a) part of this question first so that you can understand this method.



(√6) cos θ + (√10)sin θ


R=4
a=52.23

4 cos(θ − 52.23)

In the ii(a) part of this question which is (√6) cos θ + (√10)sin θ = −4, we'll be needing to modify the range first.

0 < θ < 360
0 - 52.23 < θ < 360 - 52.23
-52.23 < θ < 307.76

With the above range, our angle may fall either in the -1 quadrant (cos positive), 1st quadrant (all positive), 2nd quadrant (sin positive), 3rd quadrant ( tan positive) or the 4th quadrant (cos positive).

(√6) cos θ + (√10)sin θ = −4
4 cos(θ − 52.23) = −4
cos (θ − 52.23) = -1

Now, we need to see that in which quadrant is 'cos' negative from the range of quadrants we found above. We have the following quadrants: -1, 1, 2, 3 & 4 and the quadrants in which 'cos' is negative are the 2nd quadrant and the 3rd quadrant. In the 2nd quadrant, its going to be '180 - (cos^-1) (1)' and in the 3rd quadrant, its going to be '180 + cos (1)'. If we solve both these equation, we get the same answer.

cos (θ − 52.23) = -1
θ − 52.23 = 180
θ = 180 + 52.23
θ = 232.23

Coming to the ii(b) part of this question, we'll again be needing to modify the range.

0/2 < θ/2 < 360/2
0 < θ/2 < 180
0 - 52.23 < θ/2 - 52.23 < 180 - 52.23
-52.23 < θ/2 - 52.23 < 127.77

Now with the above range, our angle may fall either in the -1 quadrant (cos positive), 1st quadrant (all positive) or the 2nd quadrant (sin positive).

(√6) cos θ/2 + (√10)sin θ/2 = 3
4 cos(θ/2 − 52.23) = 3

Now, for this part,we need to see that in which quadrant is 'cos' positive from the range of quadrants we found above. We have the following quadrants: -1, 1 and 2 and the quadrants in which 'cos' is positive is the -1 quadrant and the 1st quadrant. In the -1 quadrant, its going to be '- (cos^-1)(3/4) ' and in the 1st quadrant, its going to be '(cos^-1)(3/4) '. In the -1 quadrant, we get θ = - 41.41 and in the 1st quadrant, we get θ = 41.41. θ = -41.41 being the smaller angle, we'll use it.


(√6) cos θ/2 + (√10)sin θ/2 = −4
4 cos(θ/2 − 52.23) = −4
cos (θ/2 − 52.23) = -1
θ/2 − 52.23 = -41.41
θ/2 = -41.41 + 52.23
θ/2 = 10.82
θ = 21.7°

okay I think I haven't been doing these kind of questions with the correct method.
For the firs part,I normally don't change the degree limits and still manage to get the correct answer but I guess your method is solely the correct one as it focuses on quadrants.I am not too good at this particular topic and will practice more tomorrow to see if I've understood it.
I've underlined the steps which I couldn't understand.Could you please explain how you got -41.41 when I understand we're supposed to find inverse of -1 but I am sure the limits have to play a part in this,just can't seem to understand how.
thanks
EDIT:I am not sure but I think you've confused the two parts.Instead of 3,you wrote -4 and hence my confusion.I now seem to have understood this topic.Just one question what are the negative quadrants?

 

[PhoenixAsh12]

I can't solve the whole question but I am sure this will help.

y=x/2 + 2
For the coordinates of A, put x=0 in the equation and you get y = 2 therefore A(0,2)
The gradient of BD = -2
D(10,-3)
-3=10(-2) + c
c=17
We got the equation of BD ;- y=-2x + 17

For the coordinates of B, set the 2 equations equal and solve.

As the question says AB=BC, it means B is the mid-point of the line.
We have the coordinates of A and B right?
xB , yB = 0 + xC /2 , 2 + yC/2
xC/2 = xB and 2+yC/2 = yB
For this part, you could have used the vector move too.

I may have missed something because I am in a hurry but the method is the same.

Thanks a tonne! I know you're in a hurry but when you're free can you please tell me what a vector move is...

 

[PhoenixAsh12]

Also, any idea about how to solve this?

 

[ffaadyy]

okay I think I haven't been doing these kind of questions with the correct method.
For the firs part,I normally don't change the degree limits and still manage to get the correct answer but I guess your method is solely the correct one as it focuses on quadrants.I am not too good at this particular topic and will practice more tomorrow to see if I've understood it.
I've underlined the steps which I couldn't understand.Could you please explain how you got -41.41 when I understand we're supposed to find inverse of -1 but I am sure the limits have to play a part in this,just can't seem to understand how.
thanks

I had accidentally mistyped that part. This is how it is:

(√6) cos θ/2 + (√10)sin θ/2 = 3
4 cos(θ/2 − 52.23) = 3
θ/2 − 52.23 = (cos^-1)(3/4)
θ/2 − 52.23 = -41.41
θ/2 = -41.41 + 52.23
θ/2 = 10.82
θ = 21.7°

We've got 41.41 from (cos^-1)(3/4). As the angle could've been in the 2 quadrants namely, the -1 quadrant and the 1st quadrant. The angle would've either been -41.41 or 41.41. We were to select the smallest possible angle therefore we chose -41.41. I hope you get it now.

 

[2pac]

I had accidentally mistyped that part. This is how it is:

(√6) cos θ/2 + (√10)sin θ/2 = 3
4 cos(θ/2 − 52.23) = 3
θ/2 − 52.23 = (cos^-1)(3/4)
θ/2 − 52.23 = -41.41
θ/2 = -41.41 + 52.23
θ/2 = 10.82
θ = 21.7°

We've got 41.41 from (cos^-1)(3/4). As the angle could've been in the 2 quadrants namely, the -1 quadrant and the 1st quadrant. The angle would've either been -41.41 or 41.41. We were to select the smallest possible angle therefore we chose -41.41. I hope you get it now.

Yup got it.Thanks a lot for your time and effort.Really appreciate it.

 

[ffaadyy]

Also, any idea about how to solve this?

What's the answer to this question?

 

[PhoenixAsh12]

What's the answer to this question?

Part (i)=25.9cm
(ii)=15.3 cm^2

 

[PhoenixAsh12]

What does it mean if 2 position vetors are perpindicular? a1b1/a2b2= -1 or what?
Please help, thanks

 

[2pac]

Also, any idea about how to solve this?

I can help u with this one.
so i part-
OPT is a triangle.You can find ot by using the pythog rule.OT will be 13cm.
perimeter of the shaded region will be 12cm+QT+arc length of OPQ.
QT will be 13-5 which is 8cm.I hope you can find the arc length using radius 5cm.Add all 3 and you'll get the perimeter.
now ii part will be easier if you've understand the concept behind the i part.
Here the area of the shaded region will be
Area of Triangle-Area of sector OPQ.
area of triangle will be 0.5XbaseXheight.
area of sector is 0.5pie r square X theeta.
theeta will be angle POT(laughs)which can be again be found by trig rule.
take any sin tan cos and you will find it.
so then subtract the two value of areas and youll get the answer.
Hope you got it.

 

[ffaadyy]

Part (i)=25.9cm
(ii)=15.3 cm^2

(i)

Find the length of the side OT using the pythagoras theorem.

OT^2 = OP^2 + PT^2
OT^2 = 5^2 + 12^2
OT^2 = 169
OT = 13

Next, we need to find the the arc length. For that, we'll first be needing to find the angle θ. Use the trigonometric property of 'tan θ = opp/adj' to find the value of θ.

tan θ = opp/adj
tan θ = 12/5
θ = 1.176

s = r θ
s = 5 x 1.176
s = 5.88cm

To find the length of the side QT, we'll simply subtract OQ (5cm) from OT (13cm).

13-5
8cm

Therefore, we now have the arc length (5.88cm), the side QT (8cm) and the side PT (12cm). Simply add all of them up to find the perimeter of the shaded area.

8 + 5.88 + 13
25.9cm

(ii)

Having found all the necessary things in the (i) part of this question, we'll only be needing to subtract the area of the sector from the area of the right angled triangle.

Area of Triangle = 0.5 x 5 x 12 = 30 cm

Area of the Sector = 0.5 x 5 x 5.88 = 14.7 cm

Area of Triangle - Area of the Sector
30 - 14.7
15.3cm

 

[GetSomeLife]

Also, any idea about how to solve this?

i) Tan OTP = 5/12
OTP = tan inverse (5/12)
OTP = 22.61986495 degrees
POQ + 90 + OTP = 180
POQ = 67.38013505

OT = 13 (by pythagoras theorem)
QT = 8

arc length = 67.38013505/360 *2 * pie * 5 = 5.88

Perimeter of shaded area = 12 + 8 + (arc length)
Perimeter of shaded area = 8 + 12 + 5.88 = 25.9

ii) Shaded area = area of triangle - area of sector = (0.5 * 12 * 5) - (67.38013505/360 * pie * 5^2)
= 30 - 14.7 = 15.3

 

[2pac]

Could someone please explain me q6(all parts) of oct 2009 p31 and also q10 of 32.I know its a lot to ask but I just can't understand Vectors.Have only understood common points of intersection so far.Would appreciate if someone could teach me the fundamentals so I could get the hang of vectors.my paper is almost 2 weeks away,so urgent help required.
thanks

 

[PhoenixAsh12]

I can help u with this one.
so i part-
OPT is a triangle.You can find ot by using the pythog rule.OT will be 13cm.
perimeter of the shaded region will be 12cm+QT+arc length of OPQ.
QT will be 13-5 which is 8cm.I hope you can find the arc length using radius 5cm.Add all 3 and you'll get the perimeter.
now ii part will be easier if you've understand the concept behind the i part.
Here the area of the shaded region will be
Area of Triangle-Area of sector OPQ.
area of triangle will be 0.5XbaseXheight.
area of sector is 0.5pie r square X theeta.
theeta will be angle POT(laughs)which can be again be found by trig rule.
take any sin tan cos and you will find it.
so then subtract the two value of areas and youll get the answer.
Hope you got it.

Thank you!

 

[PhoenixAsh12]

i) Tan OTP = 5/12
OTP = tan inverse (5/12)
OTP = 22.61986495 degrees
POQ + 90 + OTP = 180
POQ = 67.38013505

OT = 13 (by pythagoras theorem)
QT = 8

arc length = 67.38013505/360 *2 * pie * 5 = 5.88

Perimeter of shaded area = 12 + 8 + (arc length)
Perimeter of shaded area = 8 + 12 + 5.88 = 25.9

ii) Shaded area = area of triangle - area of sector = (0.5 * 12 * 5) - (67.38013505/360 * pie * 5^2)
= 30 - 14.7 = 15.3

Thanks!

 

[PhoenixAsh12]

(i)

Find the length of the side OT using the pythagoras theorem.

OT^2 = OP^2 + PT^2
OT^2 = 5^2 + 12^2
OT^2 = 169
OT = 13

Next, we need to find the the arc length. For that, we'll first be needing to find the angle θ. Use the trigonometric property of 'tan θ = opp/adj' to find the value of θ.

tan θ = opp/adj
tan θ = 12/5
θ = 1.176

s = r θ
s = 5 x 1.176
s = 5.88cm

To find the length of the side QT, we'll simply subtract OQ (5cm) from OT (13cm).

13-5
8cm

Therefore, we now have the arc length (5.88cm), the side QT (8cm) and the side PT (12cm). Simply add all of them up to find the perimeter of the shaded area.

8 + 5.88 + 13
25.9cm

(ii)

Having found all the necessary things in the (i) part of this question, we'll only be needing to subtract the area of the sector from the area of the right angled triangle.

Area of Triangle = 0.5 x 5 x 12 = 30 cm

Area of the Sector = 0.5 x 5 x 5.88 = 14.7 cm

Area of Triangle - Area of the Sector
30 - 14.7
15.3cm

Thank you! Quick question though, do i have to use radians?

 

[ffaadyy]

Thank you! Quick question though, do i have to use radians?

Yes, you've to switch your calculator to the radian mode when finding the angle of the sector or dealing with any question related to this chapter.

 

[PhoenixAsh12]

Yes, you've to switch your calculator to the radian mode when finding the angle of the sector or dealing with any question related to this chapter.

Woah never knew that, thanks!!

 

[PhoenixAsh12]

I'm sorry for bombarding the thread... My exam's after tomorrow and there's no one/place but here to help...
O/N 10 paper 12 question 5

m/j 09 paper 1 question 10

MJ 07 paper1 question 6- how to find D? I know we find the lengh of BC then equate it to something?

Please help if you have the time I would really appreciate it

 

[SUMMERLOV35]

salam can someone help with question 1
http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_w11_qp_13.pdf