Use vectors to find interior angles of the two triangles with given vertices - round to nearest hundredths
Question 1: (-3, -4),(1,7),(8,2)
Question 2: (-3,5), (-1,9), (7,9)
To find the interior angles of a triangle using vectors, we can use the dot product formula. The dot product of two vectors A and B is given by the equation:
A · B = |A| |B| cos(θ)
where |A| and |B| are the magnitudes of vectors A and B, and θ is the angle between them.
Let's solve the two questions using this method:
Question 1: (-3, -4), (1, 7), (8, 2)
First, we need to find the vectors that represent two sides of the triangle. We can find the vectors by subtracting the coordinates of the endpoints:
Vector AB = (1, 7) - (-3, -4) = (4, 11)
Vector BC = (8, 2) - (1, 7) = (7, -5)
Vector CA = (-3, -4) - (8, 2) = (-11, -6)
Now, let's find the angles. We can use the dot product formula:
Angle ABC = arccos((AB · BC) / (|AB| |BC|))
Angle BCA = arccos((BC · CA) / (|BC| |CA|))
Angle CAB = arccos((CA · AB) / (|CA| |AB|))
To calculate these angles, we need to calculate the magnitudes of the vectors:
|AB| = sqrt(4^2 + 11^2) = sqrt(16 + 121) = sqrt(137)
|BC| = sqrt(7^2 + (-5)^2) = sqrt(49 + 25) = sqrt(74)
|CA| = sqrt((-11)^2 + (-6)^2) = sqrt(121 + 36) = sqrt(157)
Now we can substitute the values into the formula:
Angle ABC = arccos((4 * 7 + 11 * (-5)) / (sqrt(137) * sqrt(74)))
Angle BCA = arccos((7 * (-11) + (-5)*(-6)) / (sqrt(74) * sqrt(157)))
Angle CAB = arccos((-11 * 4 + (-6) * 11) / (sqrt(157) * sqrt(137)))
Calculating these angles will give you the interior angles of the triangle.
Question 2: (-3, 5), (-1, 9), (7, 9)
Similarly, we find the vectors representing two sides of the triangle:
Vector AB = (-1, 9) - (-3, 5) = (2, 4)
Vector BC = (7, 9) - (-1, 9) = (8, 0)
Vector CA = (-3, 5) - (7, 9) = (-10, -4)
Now, let's find the angles using the dot product formula:
|AB| = sqrt(2^2 + 4^2) = sqrt(4 + 16) = sqrt(20)
|BC| = sqrt(8^2 + 0^2) = sqrt(64) = 8
|CA| = sqrt((-10)^2 + (-4)^2) = sqrt(100 + 16) = sqrt(116)
Angle ABC = arccos((2 * 8 + 4 * 0) / (sqrt(20) * 8))
Angle BCA = arccos((8 * (-10) + 0 * (-4)) / (8 * sqrt(116)))
Angle CAB = arccos((-10 * 2 + (-4) * 8) / (sqrt(116) * sqrt(20)))
By substituting the values and calculating, you will find the interior angles of the triangle.
all are done the same way, using the definition of the dot product:
u•v = |u|*|v|*cosθ
so, the angle between (-3, -4) and (1,7) can be figured:
-3-28 = -31 = 5*√50*cosθ
cosθ = -31/(5√50)
θ = 151.26°
proceed similarly with the others