triangle ABC vertices are A(-4,2), B(4,3), C(1,-3). list the angles in order from least measure to greatest?

Answer 1

What are the lengths of the sides?
Why would that be helpful to know?
What kind of triangle does it create?
Why would that be helpful to know?

Answer 2

(-4,2) and (4,3)
D = sqrt((4 - (-4))^2 + (3 - 2)^2)
D = sqrt((4 + 4)^2 + 1^2)
D = sqrt(8^2 + 1)
D = sqrt(64 + 1)
D = sqrt(65)

(4,3) and (1,-3)
D = sqrt((1 - 4)^2 + (-3 - 3)^2)
D = sqrt((-3)^2 + (-3 + (-3))^2)
D = sqrt(9 + (-6)^2)
D = sqrt(9 + 36)
D = sqrt(45)

(-4,2) and (1,-3)
D = sqrt((1 - (-4))^2 + (-3 - 2)^2)
D = sqrt((1 + 4)^2 + (-3 + (-2))^2)
D = sqrt(5^2 + (-5)^2)
D = sqrt(25 + 25)
D = sqrt(50)

c^2 = a^2 + b^2 - 2ab(cos(C))

sqrt(65)^2 = sqrt(45)^2 + sqrt(50)^2 - 2(sqrt(45)sqrt(50))cos(C)
65 = 45 + 50 - 2sqrt(2250)cos(C)
65 = 95 - 2sqrt(225 * 10)cos(C)
-30 = -30sqrt(10)cos(C)
1 = sqrt(10)cosC
cosC = 1/(sqrt(10))
cosC = (sqrt(10))/10
C = about 71.565°

a^2 = b^2 + c^2 - 2bc(cosA)
sqrt(45)^2 = sqrt(50)^2 + sqrt(65)^2 - 2(sqrt(65) * sqrt(50))cosA
45 = 50 + 65 - 2sqrt(3250)cosA
45 = 115 - 2sqrt(25 * 130)cosA
-70 = -10sqrt(130)cosA
7 = sqrt(130)cosA
cosA = 7/(sqrt(130))
cosA = (7sqrt(130))/130
A = about 52.125°

b^2 = a^2 + c^2 - 2ac(cosB)

sqrt(50)^2 = sqrt(45)^2 + sqrt(65)^2 - 2(sqrt(45)sqrt(65))cosB
50 = 45 + 65 - 2sqrt(2925)cosB
50 = 110 - 2sqrt(225 * 13)cosB
-60 = -30sqrt(13)cosB
2 = sqrt(13)cosB
cosB = 2/sqrt(13)
cosB = (2sqrt(13))/13
B = about 56.3°

ANS : To the nearest Degree

52°, 56°, 72°

if you add the actual values together, you will still get 180.

Answer 3

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Answer 4

Use distance formula with your coordinates and trig functions with the lengths of the sides or law of sines.

Answer 5

The sides of the triangle are:

AB = sqrt(65)
BC = sqrt(45)
CA = sqrt(50)

From the law of cosines you get that:

AB^2 = BC^2 + CA^2 - 2*BC*CA*cos(ACB)
BC^2 = CA^2 + AB^2 - 2*CA*AB*cos(BAC)
CA^2 = AB^2 + BC^2 - 2*AB*BC*cos(CBA)

then

65 = 45 + 50 - 2*sqrt(45*50)*cos(ACB)
45 = 50 + 65 - 2*sqrt(50*65)*cos(BAC)
50 = 65 + 45 - 2*sqrt(65*45)*cos(CBA)

and

Angle ACB = 71.565 degrees
Angle BAC = 52.125 degrees
Angle CBA = 56.310 degrees

I hope this helps.

Answer 6

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