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
If $\displaystyle P,~Q,~\&~R$ are three points, then the angle between $\displaystyle \overrightarrow {PQ} \;\& \,\overrightarrow {PR} $ is $\displaystyle \arccos \left( {\frac{{\overrightarrow {PQ} \cdot \overrightarrow {PR} }}{{\left\| {\overrightarrow {PQ} } \right\|\left\| {\overrightarrow {PR} } \right\|}}} \right)$
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 "$\displaystyle \phi$" 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 $\displaystyle \phi$, so the slope of the line is tan(\phi)= tan(a2- a1).
There is an identity that says that $\displaystyle tan(a2- a1)= \frac{tan(a2)- tan(a1)}{1+ tan(a1)tan(a2)}$ (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.
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
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
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