To determine the size of angle BAC and the area ratio of triangle ABC to triangle EDC, we can follow these steps:
**Step 1: Calculate the Slopes of Lines AB and AE:**
The slope (m) of a line passing through two points (x1, y1) and (x2, y2) is given by:
m = (y2 - y1) / (x2 - x1)
For line AB:
- A(6, 16)
- B(11, 12)
m_AB = (12 - 16) / (11 - 6) = -4 / 5
For line AE:
- A(6, 16)
- E(6, -1)
m_AE = (-1 - 16) / (6 - 6) = undefined (vertical line)
**Step 2: Determine Angle BAC:**
The angle BAC is formed between lines AB and AE. To find the angle between two lines, you can use the tangent of the angle (tan θ), which is equal to the absolute value of the difference in their slopes divided by 1 plus the product of their slopes:
tan θ = |(m_AB - m_AE) / (1 + m_AB * m_AE)|
tan θ = |(-4/5 - undefined) / (1 + (-4/5) * undefined)|
tan θ = |(-4/5) / (1)|
tan θ = 4/5
Now, find θ by taking the arctan (inverse tangent) of 4/5 using a calculator:
θ = arctan(4/5)
θ ≈ 38.66 degrees
So, the size of angle BAC is approximately 38.66 degrees.
**Step 3: Determine the Area Ratio of Triangle ABC to Triangle EDC:**
To find the area ratio of triangle ABC to triangle EDC, we need to calculate the areas of both triangles.
Area of Triangle ABC = (1/2) * AB * AC
Area of Triangle EDC = (1/2) * ED * EC
First, find the lengths of AC and EC:
- A(6, 16)
- C(6, K)
AC = |16 - K| (vertical line)
- E(6, -1)
- C(6, K)
EC = |-1 - K| (vertical line)
Now, calculate the areas:
Area of Triangle ABC = (1/2) * AB * AC = (1/2) * 5 * |16 - K| = 5 * |16 - K| / 2
Area of Triangle EDC = (1/2) * ED * EC = (1/2) * 10 * |-1 - K| = 10 * |-1 - K| / 2
Now, find the ratio of the areas:
Area Ratio = (Area of Triangle ABC) / (Area of Triangle EDC) = (5 * |16 - K| / 2) / (10 * |-1 - K| / 2)
Area Ratio = (5 * |16 - K|) / (10 * |-1 - K|)
Simplify by canceling out the common factors:
Area Ratio = (5 * |16 - K|) / (10 * |-1 - K|)
Area Ratio = (|16 - K|) / (2 * |-1 - K|)
The area ratio is given by this expression, and you can further simplify it if needed.