Three deer, A, B, and C, are grazing in a field. Deer B is located 61.0 m from deer A at an angle of 54.1 ° north of west. Deer C is located 79.8 ° north of east relative to deer A. The distance between deer B and C is 98.1 m. What is the distance between deer A and C?
figure the angle θ between the lines AB and BC. Then use the law of cosines to get AC.
AC^2 = 61^2 + 98.1^2 - 2(61)(98.1)cosθ
Or, you can use the law of sines twice to find angle C and then AC.
To find the distance between deer A and C, we need to use the concept of vector addition.
1. First, let's convert the angles given to the standard mathematical coordinate system (where the positive x-axis points east and the positive y-axis points north).
- Angle north of west: 54.1°
This angle is measured clockwise from the positive y-axis, meaning it's 90° - 54.1° = 35.9° north of the positive y-axis.
- Angle north of east: 79.8°
This angle is measured counterclockwise from the positive x-axis, meaning it's 180° - 79.8° = 100.2° north of the positive y-axis.
2. Next, we represent the displacement vectors for deer B and C:
- Deer B:
- Magnitude: 61.0 m
- Angle: -35.9° (note the negative sign, indicating it's south of the positive y-axis)
- Deer C:
- Magnitude: 98.1 m
- Angle: 100.2°
3. We can now resolve the displacement vectors into their x and y components:
- For deer B:
- x-component: -61.0 m × sin(35.9°) ≈ -35.97 m (east direction is positive)
- y-component: 61.0 m × cos(35.9°) ≈ 50.62 m (north direction is positive)
- For deer C:
- x-component: 98.1 m × cos(100.2°) ≈ -15.84 m (east direction is positive)
- y-component: 98.1 m × sin(100.2°) ≈ 95.53 m (north direction is positive)
4. Now, we can find the total displacement of deer A relative to deer C by adding the corresponding x and y components:
- x-component total: -35.97 m - (-15.84 m) = -20.13 m (east direction is positive)
- y-component total: 50.62 m + 95.53 m = 146.15 m (north direction is positive)
5. Finally, we can calculate the magnitude of the total displacement vector using the Pythagorean theorem:
- Distance between deer A and C:
= √((-20.13 m)² + (146.15 m)²)
≈ √(406.1996 + 21360.5225)
≈ √21766.7221
≈ 147.54 m
Therefore, the distance between deer A and C is approximately 147.54 m.
To find the distance between deer A and C, we can use the Law of Cosines.
First, let's label the given information:
AB = 61.0 m (distance between deer A and B)
angle ABD = 54.1° (angle north of west)
angle ABC = 180° - angle ABD = 180° - 54.1° = 125.9°
AC = ? (distance between deer A and C)
angle ACB = 79.8° (angle north of east)
BC = 98.1 m (distance between deer B and C)
Now, we can apply the Law of Cosines, which states:
c^2 = a^2 + b^2 - 2ab * cos(C)
where:
- c is the unknown side (AC)
- a and b are the known sides (AB and BC)
- C is the angle opposite to the unknown side (angle ABC)
Substituting the values into the formula, we get:
AC^2 = AB^2 + BC^2 - 2 * AB * BC * cos(angle ABC)
AC^2 = 61.0^2 + 98.1^2 - 2 * 61.0 * 98.1 * cos(125.9°)
Calculating this expression:
AC^2 = 3721.0 + 9622.1 - 2 * 61.0 * 98.1 * (-0.5736)
AC^2 = 3721.0 + 9622.1 + 336_535.56
AC^2 ≈ 348_878.66
Taking the square root of both sides to solve for AC:
AC ≈ sqrt(348_878.66)
AC ≈ 590.68
Therefore, the distance between deer A and C is approximately 590.68 meters.