831.Let A = (1,3), B = (6,0), and C = (9,9). Find the size of angle BAC.
figure the lengths of the sides. Then use the law of cosines.
c^2 = a^2 + b^2 - 2ab cosC
To find the size of angle BAC, we need to find the slopes of AB and AC first.
The slope of AB is calculated as:
m_AB = (y2 - y1) / (x2 - x1)
Plugging in the coordinates of A and B, we get:
m_AB = (0 - 3) / (6 - 1) = -3/5
Similarly, the slope of AC is calculated as:
m_AC = (y3 - y1) / (x3 - x1)
Plugging in the coordinates of A and C, we get:
m_AC = (9 - 3) / (9 - 1) = 6/8 = 3/4
Now, we can find the angle BAC using the formula:
tan(angle BAC) = |(m_AB - m_AC) / (1 + m_AB * m_AC)|
Substituting the values:
tan(angle BAC) = |((-3/5) - (3/4)) / (1 + (-3/5) * (3/4))|
tan(angle BAC) = |((-12/20) - (15/20)) / (1 + (-9/20))|
tan(angle BAC) = |((-27/20) / (11/20))|
tan(angle BAC) = |(-27/11)|
We can now use an inverse tangent function to find the angle BAC:
angle BAC = atan(|(-27/11)|)
Calculating this, we find:
angle BAC ≈ 1.176 radians or ≈ 67.38 degrees (rounded to two decimal places)
Therefore, the size of angle BAC is approximately 67.38 degrees.
To find the size of angle BAC, we can make use of the concept of slope and the dot product between vectors.
Step 1: Find the slope of AB and AC.
The slope of a line passing through two points (x1, y1) and (x2, y2) is given by the formula:
slope = (y2 - y1) / (x2 - x1)
Using the coordinates of A, B, and C, we can find the slopes of AB and AC:
Slope of AB = (0 - 3) / (6 - 1) = -3/5
Slope of AC = (9 - 3) / (9 - 1) = 6/8 = 3/4
Step 2: Use the dot product to find the angle.
The dot product of two vectors is defined as:
dot product = |A| * |B| * cos(theta)
Here, |A| and |B| represent the magnitudes of vectors A and B, and theta represents the angle between them.
Using the slopes of AB and AC, we can calculate the angle BAC as follows:
cos(theta) = (dot product) / (|AB| * |AC|)
= (AB dot AC) / (|AB| * |AC|)
For convenience, we can calculate the dot product and magnitudes of vectors AB and AC:
AB dot AC = (slope of AB) * (slope of AC)
= (-3/5) * (3/4)
= -9/20
|AB| = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((6 - 1)^2 + (0 - 3)^2)
= sqrt(25 + 9)
= sqrt(34)
|AC| = sqrt((x2 - x1)^2 + (y2 - y1)^2)
= sqrt((9 - 1)^2 + (9 - 3)^2)
= sqrt(64 + 36)
= sqrt(100)
= 10
Substituting these values in the equation cos(theta) = (AB dot AC) / (|AB| * |AC|), we get:
cos(theta) = (-9/20) / (sqrt(34) * 10)
= (-9/20) / (10 sqrt(34))
= -9 / (20 * 10 sqrt(34))
= -9 / (200 sqrt(34))
Step 3: Calculate the angle BAC.
We have the value of cos(theta), but to find theta itself, we use the inverse cosine function (arccos).
theta = arccos(cos(theta))
= arccos(-9 / (200 sqrt(34)))
Using a calculator, we find that arccos(-9 / (200 sqrt(34))) is approximately 51.07 degrees.
Therefore, the size of angle BAC is approximately 51.07 degrees.