Angle between tangents/line. Math P1.

[SZ11]

I need to know how to calculate the acute/obtuse angles between tangents to a curve or between a tangent and a line. Need urgent help with this.
Questions like this can be found in past papers. N05Q9 And N08Q9. Help Help Help!!! :S

 

[limca22334]

well this is a direct ans to the nov05 q9:
well 1) gradient of a line is also defined as tan of the angle tht line makes with the +ve dir of the x-axis
2) if gradient is +ve angle made is acute and if -ve angle obtuse
3) in this Q both the tangents and the linel have - ve gradients.......

now this is a little hard to explain as i cannot draw..................here goes;
gradient of line=-2..........tan-1(inverse)-2= 63.4
180-63.4 (cuz u need obtuse angle)= 116.6

perform similar procedure for the curve/............gradient= -3............(shift tan) tan-1[inverse]-3=71.6 -------> 190-71.6= 108.4

116.6-108.4-8.2<----ANSWER as all angles in a triangle are 18- deg

REALLY HOPE THIS HELPS GOOD LUCK FOR 2MOROW AND IF SOME1 HAS A BETTER WAY.....PLZ EXPLAIN THNX
salam

 

[SZ11]

Thanks man!
I realised the formula was tanQ = gradient.
Now I know it's the same for every question about angles.
Thanks again!

 

[1992]

how do u do the question before it?? Q5.ii)

 

[1992]

I still dont understand it. what do u mean of using tangent?? :S

 

[Zazzyo]

I used this. tan(x)=tan(M)-tan(m)/ 1 + tan(M)tan(m)

where M and m are gradients of two lines. Its a p3 formula but hassle free if u memorise it.

 

[coptir]

There is a simpler formula for this:
tanX= (m1-m2)/(1+m1m2)

where m1 is the gradient of first line and m2 is the gradient of the second line and X=theta

 

[1992]

could someone explain this Q.i) b and c from O/N 2009 /12
Its seem that this Question are coming up in recent pastpapers. need help pleaseee!!

 

[1992]

There is a simpler formula for this:
tanX= (m1-m2)/(1+m1m2)

where m1 is the gradient of first line and m2 is the gradient of the second line and X=theta

Do u know anywhere which can explain this formula?? as i have never seen it before. it seems new this types of questions :SS
Tomorrow is exam.

or could u elaborate more to how i can solve this question with that formula

 

[PlanetMaster]

Do u know anywhere which can explain this formula?? as i have never seen it before. it seems new this types of questions :SS
Tomorrow is exam.

or could u elaborate more to how i can solve this question with that formula

I'll assume that you know the equations of the two lines, from which you can get their slopes.
Suppose they are m1 and m2. Since the slope of a line is the tangent of its angle of elevation,
the angle between the lines is the difference in the two angles.
Thus, if theta is the angle between, and A and B are the angles of elevation of the two lines,


Tan(theta) = Tan(A-B) = .Tan(A) - Tan(B). = ..m1 - m2..
Tan(theta) = Tan(A-B) = 1 + Tan(A)Tan(B)....1 + m1*m2

Here I used the identity:
Tan(x-y) = [Tan(x) - Tan]/[1 + Tan(x)Tan]

So the angle you are looking for is the inverse tangent of
(m1 - m2)/(1 + m1*m2).

 

[1992]

so do i need to calculate the gradient for the 2 lines. and after that how i can calculate the angle between them¿¿

 

[1992]

so for these types of questions we just need to calculate the gradient and use the identity???

 

[XWTBSILT]

Its easier if use vectors then no need of subtraction u straight away get the correct answer

 

[Anomaly]

The gradient is also defined as Tan (Theta), thus you should find the gradient at a point on the curve and find tan inverse of that gradient. The angel you get is the angel made by the tangent (at that point on the curve) with the x axis, it helps to extend the line to the axis and draw the angel. For a line do the same thing, re-arrange the eq. to find gradient and find the angle. After drawing the angels, they usually form a triangle from which you can calculate the original angle b/w line and curve