Find, correct to the nearest degree, the three angles of the triangle with the given vertices.?
A(2, 0),B(4, 5), C(-3,2)
ANGLE- CAB
ANGLE- ABC
ANGLE- BCA
In degrees.
I got the vectors AB <,5>
BC <1,3>
AC <-1,2>
and I know the equation b.c= |b| |c| cos t
I have no idea how you were taught to calculate vectors.
For any two points P(a,b) and Q(c,d)
vector PQ = < c-a, d-b>
I got
vector AB = <2,5>
vector BC = <-7,-3) and
vector AC = <-5,2>
let angle CAB = Ø
then vector AC . vector AB = |AC| |AB| cosØ
<-5,2> . <2,5> = | <-5,2>| |<2,5>| cosØ
-10+10 = √29 √29 cosØ
cosØ = 0
Ø = 90°
well, that was obvious from looking at the two vectors, their dot product is zero, so they are perpendicular
for angle ABC
<-3,-7> . <-5,-2> = √58 √29 cos B
15+14 = √58√29cosB
cos B =.707106...
B = 45°
Well, how about that ?
Suppose we had taken the length of each vector
|AB = √29
|AC| = √29
|BC| = √58
so it is isosceles, and since
(√58) ^2 = (√29)^2 + (√29)^2
so it is also right-angled, as I showed in my first method.
I have no idea how you were taught to calculate vectors.
For any two points P(a,b) and Q(c,d)
vector PQ = < c-a, d-b>
I got
vector AB = <2,5>
vector BC = <-7,-3) and
vector AC = <-5,2>
let angle CAB = Ø
then vector AC . vector AB = |AC| |AB| cosØ
<-5,2> . <2,5> = | <-5,2>| |<2,5>| cosØ
-10+10 = √29 √29 cosØ
cosØ = 0
Ø = 90°
well, that was obvious from looking at the two vectors, their dot product is zero, so they are perpendicular
for angle ABC
<-3,-7> . <-5,-2> = √58 √29 cos B
15+14 = √58√29cosB
cos B =.707106...
B = 45°
Well, how about that ?
Suppose we had taken the length of each vector
|AB = √29
|AC| = √29
|BC| = √58
so it is isosceles, and since
(√58) ^2 = (√29)^2 + (√29)^2
so it is also right-angled, as I showed in my first method.
THANK YOU SO MUCH!
To find the angles of the triangle with the given vertices, we can use the dot product of vectors.
First, let's find the vectors AB, BC, and AC:
AB = B - A = (4,5) - (2,0) = (2,5)
BC = C - B = (-3,2) - (4,5) = (-7,-3)
AC = C - A = (-3,2) - (2,0) = (-5,2)
Next, we need to calculate the magnitudes of vectors AB, BC, and AC:
|AB| = √(2^2 + 5^2) = √(4 + 25) = √29
|BC| = √((-7)^2 + (-3)^2) = √(49 + 9) = √58
|AC| = √((-5)^2 + 2^2) = √(25 + 4) = √29
Now, we can calculate the cosines of the angles using the dot product formula:
cos(CAB) = (AB · AC) / (|AB| |AC|)
cos(ABC) = (AB · BC) / (|AB| |BC|)
cos(BCA) = (BC · AC) / (|BC| |AC|)
Let's calculate each angle:
cos(CAB) = (AB · AC) / (|AB| |AC|)
= (2 * -5 + 5 * 2) / (√29 * √29)
= (0) / 29
= 0
cos(ABC) = (AB · BC) / (|AB| |BC|)
= (2 * -7 + 5 * -3) / (√29 * √58)
= (-14 - 15) / (√29 * √58)
= -29 / (√29 * √58)
= -1 / √58
cos(BCA) = (BC · AC) / (|BC| |AC|)
= (-7 * -5 + -3 * 2) / (√58 * √29)
= (35 - 6) / (√58 * √29)
= 29 / (√58 * √29)
= 1 / √2
To find the angles, we need to take the arccosine of each of the cosines:
angle CAB ≈ arccos(cos(CAB)) ≈ arccos(0) ≈ 90 degrees
angle ABC ≈ arccos(cos(ABC)) ≈ arccos(-1 / √58) ≈ 131.81 degrees (rounded to the nearest degree)
angle BCA ≈ arccos(cos(BCA)) ≈ arccos(1 / √2) ≈ 45 degrees (rounded to the nearest degree)
Therefore, the three angles of the triangle with the given vertices are approximately:
ANGLE CAB = 90 degrees
ANGLE ABC = 132 degrees
ANGLE BCA = 45 degrees