Given the points A(0,0), B(4,3) and C(2,9), what is the measure of <ABC?
figure side lengths, then use the law of cosines.
AB^2 = 25
BC^2 = 40
AC^2 = 85
AC^2 = AB^2 + BC^2 - 2*AB*BC*cosABC
85 = 25+40-2√1000 * cosABC
<ABC =~ 109°
Thanks Steve
To find the measure of angle ABC, we can use the formula for the angle between two vectors. Here's the step-by-step process:
1. Calculate the vectors AB and BC.
- Vector AB = (x2 - x1, y2 - y1) = (4 - 0, 3 - 0) = (4, 3)
- Vector BC = (x3 - x2, y3 - y2) = (2 - 4, 9 - 3) = (-2, 6)
2. Calculate the dot product of AB and BC.
- Dot product = (AB) · (BC) = (4)(-2) + (3)(6) = -8 + 18 = 10
3. Find the magnitudes of vectors AB and BC.
- Magnitude of AB = √(4^2 + 3^2) = √(16 + 9) = √25 = 5
- Magnitude of BC = √((-2)^2 + 6^2) = √(4 + 36) = √40 = 2√10
4. Apply the dot product formula to find the cosine of the angle between AB and BC.
- Cosineθ = (AB · BC) / (|AB| * |BC|) = 10 / (5 * 2√10)
5. Simplify the expression.
- Cosineθ = 10 / (10√10) = 1 / √10
6. Use an inverse cosine function to find the angle measure.
- θ = arccos(1 / √10)
7. Calculate the angle measure using a calculator.
- θ ≈ 22.62 degrees (rounded to two decimal places)
Therefore, the measure of angle ABC is approximately 22.62 degrees.