Mathematics: Post your doubts here!

[iKhaled]

hey wat was the frmula of finding acute angel between two tangents?

tan theta = tangent

is that what you r lookin for ?

 

[Rutzaba]

i got wat i was looking for but sti dunno how to type it
http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

9 iii

 

[PANDA-]

i got wat i was looking for but sti dunno how to type it
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

9 iii

a^2 = b^2 + c^2 -2abcosq ... I guess? Lol
Would take some time because you'd have to find the point of intersection and then find the two lengths of the tangents from their respective points of tangency to the point of intersection (i hope this part is clear ) and find the length of c, the line between q and p... and then substitute.

Obviously there must be a shorter simpler way.

 

[Rutzaba]

i got wat i was looking for but sti dunno how to type it
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

9 iii

anyone helping?

 

[Rutzaba]

a^2 = b^2 + c^2 -2abcosq ... I guess? Lol
Would take some time because you'd have to find the point of intersection and then find the two lengths of the tangents from their respective points of tangency to the point of intersection (i hope this part is clear ) and find the length of c, the line between q and p... and then substitute.

Obviously there must be a shorter simpler way.

u guess wrong
tan theta= grad of one tangent
tan theta = gradient of second tangent
theta 1- theta 2 = answer
i dunno how to make the diagram is wayyy to complicated

 

[iKhaled]

i got wat i was looking for but sti dunno how to type it
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_1.pdf

9 iii

see..

dy/dx = tangent
tangent = y/x which is same as tan theta = y/x

so tan theta = dy/dx....find the gradient at point p and q by substituting the x values of p and q in the dy/dx u got (if u don't know how to differentiate tell me)

tan theta = 1.5 at point p and tan theta = 0.75 at point q
theta = 56.3 and 36.87
find the difference between the 2 56.3 - 36.87 = 19.4 and thats out answer!

i hope you got it

 

[Rutzaba]

see..

dy/dx = tangent
tangent = y/x which is same as tan theta = y/x

so tan theta = dy/dx....find the gradient at point p and q by substituting the x values of p and q in the dy/dx u got (if u don't know how to differentiate tell me)

tan theta = 1.5 at point p and tan theta = 0.75 at point q
theta = 56.3 and 36.87
find the difference between the 2 56.3 - 36.87 = 19.4 and thats out answer!

i hope you got it

wat no diagrams? ok heres one khalid to another MKh@lid

 

[iKhaled]

wat no diagrams? ok heres one khalid to another @mkh@MKh@lid

why do u need diagrams? and what is " @mkh@MKh@lid" :O:O

 

[Rutzaba]

why do u need diagrams? and what is " @mkh@MKh@lid" :O:O

see the edited version dude

 

[PANDA-]

u guess wrong
tan theta= grad of one tangent
tan theta = gradient of second tangent
theta 1- theta 2 = answer
i dunno how to make the diagram is wayyy to complicated

Take a look at iKhaled's solution. Although mine takes a way different route, by drawing the tangents and completing traingle QPI, where I is the point of intersection of the two tangents. Length of one of the tangents would be a, the other would be b, and c would be length QP. Then substitute it all in a^2 = b^2 + c^2 -2abcosq.

 

[Rutzaba]

Take a look at iKhaled's solution. Although mine takes a way different route, by drawing the tangents and completing traingle QPI, where I is the point of intersection of the two tangents. Length of one of the tangents would be a, the other would be b, and c would be length QP. Then substitute it all in a^2 = b^2 + c^2 -2abcosq.

y use cos rule? wen there is already a formula there... the ans myt have been ryt but this is way easier

 

[PANDA-]

y use cos rule? wen there is already a formula there... the ans myt have been ryt but this is way easier

I realize there's an easier way. But I don't know what exactly is the easier way lol... Care to explain it?

 

[Rutzaba]

I realize there's an easier way. But I don't know what exactly is the easier way lol... Care to explain it?

there are two lines ryt? keep saying yes or no

 

[PhyZac]

So in this case its 1.966 radians => approx. 2/3π ? We just estimate the point on the graph right?

Yes very correct, just estimate. It is only sketch, you dont want to waste time.

 

[PANDA-]

there are two lines ryt? keep saying yes or no

Yes... get to the point please. Rofl I'm not an idiot, I can get it in one go.

 

[Rutzaba]

well we did it in one go if u see the previous post but since u dint get it ... ah well
we have two lines thus two gradients ryt?
so put in formula tan x = gradieant of line 1
tan y= gradient of line two
y-x will give you the answer kapish?

 

[Rutzaba]

and since angles are never negative...

 

[champ-student]

i have to find 3 values of x and y to plot a graph...but they shud be in whole number.
11 x + 16y= 176

 

[PANDA-]

well we did it in one go if u see the previous post but since u dint get it ... ah well
we have two lines thus two gradients ryt?
so put in formula tan x = gradieant of line 1
tan y= gradient of line two
y-x will give you the answer kapish?

tbh I didn't read it, it was too long
This is much shorter... thanks.
and btw it's written "capiche".

 

[Rutzaba]

tbh I didn't read it, it was too long
This is much shorter... thanks.
and btw it's written "capiche".

thankyou for increasing my knowledge buddy
and any tym