Mathematics: Post your doubts here!

[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

 

[PhyZac]

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

(16,0)
(0,11)

These are two.

 

[PANDA-]

(16,0)
(0,11)

These are two.

That's enough really because it's a straight line lol, no use of a third point

 

[Rutzaba]

(16,0)
(0,11)

These are two.

wat is this p1?

 

[syed1995]

anyone helping?

Well knowing you.. you would have gotten this figured by now .. but here's the solution anyway..

You will find the gradients of the tangents to the curve..

m of tangents = 1.5 and 0.75

then tan-1(gradient 1) - tan-1(gradient 2)
tan^-1 (1.5) - tan^-1(0.75)
56.3 - 36.9
= 19.4 Degrees

 

[champ-student]

(16,0)
(0,11)

These are two.

oooo....i have watd so much tym wid my calculator...thx a lot..^_^

That's enough really because it's a straight line lol, no use of a third point

thx ...thx...ty sooooo much

 

[PhyZac]

wat is this p1?

Well I don't know. But it can be anything.

 

[Rutzaba]

Well knowing you.. you would have gotten this figured by now .. but here's the solution anyway..

You will find the gradients of the tangents to the curve..

m of tangents = 1.5 and 0.75

then tan-1(gradient 1) - tan-1(gradient 2)
tan^-1 (1.5) - tan^-1(0.75)
56.3 - 36.9
= 19.4 Degrees

dude.. and guys at random... post krne se pehle dekkh liya karo ke kisi aur ne swal ka jawab tou nhi dde dya.. besides i was asking fr sumone else... thnkoo anyways buddy

 

[Rutzaba]

Well I don't know. But it can be anything.

lawl then how did u get the answer

 

[Rutzaba]

Well knowing you.. you would have gotten this figured by now .. but here's the solution anyway..

thnks fr da compliments