what is the angle of A , B and C?

Given:

A(-3,-1) , B (6,3) , C (4,-2)

is this the formula? and teach me how to solve.

tanΦ=tan(α2-α1)

tanΦ=tanα2-tanα1/1+tanα2tanα1

tanΦ=m2-m1/1+m2m1

Thanks :)

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- Jan 7th 2013, 06:06 PMninearlIntersection Between Lines
what is the angle of A , B and C?

Given:

A(-3,-1) , B (6,3) , C (4,-2)

is this the formula? and teach me how to solve.

tanΦ=tan(α2-α1)

tanΦ=tanα2-tanα1/1+tanα2tanα1

tanΦ=m2-m1/1+m2m1

Thanks :) - Jan 7th 2013, 06:49 PMPlatoRe: Intersection Between Lines
- Jan 7th 2013, 07:07 PMninearlRe: Intersection Between Lines
can you solve it? i can't understand the formula hehe.. thankz

:) - Jan 7th 2013, 07:14 PMHallsofIvyRe: Intersection Between Lines
First, I take it that by "the angle of A , B and C", you mean the angle with vertex B, A lying on one ray, C on the other. What you have written further

"tanΦ=tan(α2-α1)

tanΦ=tanα2-tanα1/1+tanα2tanα1

tanΦ=m2-m1/1+m2m1"

makes no sense because you have not said what "a2", "a1", "m2", m1, and " " mean.

What**is**true is this. If ray BA makes angle a1 with the x-axis (or any horizontal line) and m1 is the slope of the line, then, by definition of "slope", m1= tan(a1). If ray BC makes angle a2 with the x-axis (or any horizontal line) and m2 is the slope of the line, then, by definition of "slope", m2= tan(a2).

Finally,**if**a2> a1, then the angle between ray BA and ray BC is a2- a1, which is what I presume you meant by , so the slope of the line is tan(\phi)= tan(a2- a1).

There is an identity that says that (which you could derive from the similar identities for the sine and cosine: sin(a2- a1)= sin(a2)cos(a1)- cos(a2)sin(a1) and cos(a2- a1)= cos(a2)cos(a1)- sin(a2)sin(a1)). Replace tan(a1) and tan(a2) in that with m1 and m2 respectively. - Jan 8th 2013, 01:45 PMbjhopperRe: Intersection Between Lines
Hi newbie,

I hope the following will help.

Plot the three points.Connect them to form triangle ABC.Draw ahorizontal line thru A and a vertical thru B meeting at D. Triangle ABD is theslope diagram for line AB. Slope = 4/9 arc tan 4/9 = 24 deg.= angleBAD

Draw a horizontal line thru C and a vertical line thru A meeting at E.Triangle ACE is the slope diagram for line AC. Slope =-1/7. Arc tangent 1/7 = 8.1 deg .= angle ACE=angle DAC. Angle A = 24 + 8.1=32.1 deg.

Do the same for angle B or C. Find third angle by sum of the three angles - Jan 9th 2013, 05:36 AMninearlRe: Intersection Between Lines
thanks alot guys :)

this would help me alot! :) - Jan 9th 2013, 07:24 AMninearlRe: Intersection Between Lines
hi bjhopper

i hope you can solve this :(

from angle A=24+8.1=32.1

from BAD arc tan 2/3 = 33.7 and arc tan 1/10 = 5.71

angle B=5.71+33.7=39.41

from CBD arc tan 5/10= 26.57 and arc tan 1/1 = 45

angle C=26.57+45=71.57

ABC=143.08

if the sum of ABC = 180 degrees it would be the correct answer right?

please help to solve this thanks :( - Jan 9th 2013, 12:18 PMbjhopperRe: Intersection Between Lines
Hi ninearl,

Did you graph and label points as described? You must do it.

Angle A is 32.1deg and angle ACE = 8.1 deg as previously described

Draw a vertical line thru B meeting a horizontal line thru C at F

Triangle BCF is the slope diagram of BC

slope of BC is 5/2 arc tan 5/2 = 68.2 deg =angle BCF

Angle ACB = 180 -8.1-68.2=103.7 deg.This is angleC

Angle B =180-103.7-32.1=44.2