A girl walks 7km on a bearing of 030° she then changes direction and walks 8km on a bearing of 120° find the distance and bearing from her starting point
Pls solve it
Did you make a diagram of the triangle?
I have mine showing sides of 7 and 8 with a 90° angle between them.
So it is just using Pythagoras
d^2 = 7^2 + 8^2 = 113
d = √113 = appr 10.63 km
For the angle of the triangle at the starting point:
tanθ = 8/7 = ....
θ = appr 48.8°
bearing from start = 48.8 + 30° = 78.8°
To find the distance and bearing from the girl's starting point, we will use the concepts of vector addition and bearing calculations.
First, let's break down the girl's movements into two separate vectors:
1. The girl walks 7km on a bearing of 030°.
To represent this movement, we can use the vector V1 = (7 km, 030°).
2. The girl then changes direction and walks 8km on a bearing of 120°.
To represent this movement, we can use the vector V2 = (8 km, 120°).
Now, let's calculate the resultant vector R, which represents the total distance and direction from the girl's starting point:
R = V1 + V2
To calculate the resultant vector, we need to break down the vectors into their vertical (North-South) and horizontal (East-West) components:
For vector V1:
Vertical Component (V1y) = 7 km * sin(30°) = 7 km * 0.5 = 3.5 km
Horizontal Component (V1x) = 7 km * cos(30°) = 7 km * 0.866 = 6.062 km
For vector V2:
Vertical Component (V2y) = 8 km * sin(60°) = 8 km * 0.866 = 6.928 km
Horizontal Component (V2x) = 8 km * cos(60°) = 8 km * 0.5 = 4 km
Now, let's calculate the resultant vector R by adding the vertical and horizontal components:
Vertical Component (Ry) = V1y + V2y = 3.5 km + 6.928 km = 10.428 km
Horizontal Component (Rx) = V1x + V2x = 6.062 km + 4 km = 10.062 km
Using the vertical and horizontal components, we can calculate the magnitude (distance) of the resultant vector R:
Magnitude of R = sqrt(Ry^2 + Rx^2) = sqrt(10.428 km^2 + 10.062 km^2) = 14.499 km (rounded to three decimal places)
To find the bearing of the resultant vector R, we can use the inverse tangent function:
Bearing of R = atan(Ry / Rx) = atan(10.428 km / 10.062 km) = atan(1.033) ≈ 47.499° (rounded to three decimal places)
Therefore, the distance from the girl's starting point is approximately 14.499 km, and the bearing is approximately 47.499°.
To find the distance and bearing from the girl's starting point, we can use trigonometry and vector addition.
1. Start by drawing a diagram to visualize the situation. Mark the girl's starting point as "A" and draw a line segment representing her first walk of 7km on a bearing of 030°. Then, from the endpoint of the first walk, draw another line segment representing her second walk of 8km on a bearing of 120°.
2. Now, let's break down each walk into their respective horizontal (x) and vertical (y) components.
- For the first walk of 7km on a bearing of 030°:
- The horizontal component (x₁) can be found using cos(30°):
x₁ = 7 km * cos(30°) ≈ 6.062 km
- The vertical component (y₁) can be found using sin(30°):
y₁ = 7 km * sin(30°) ≈ 3.5 km
- For the second walk of 8km on a bearing of 120°:
- The horizontal component (x₂) can be found using sin(30°):
x₂ = 8 km * sin(120°) ≈ 6.928 km
- The vertical component (y₂) can be found using cos(30°):
y₂ = 8 km * cos(120°) ≈ -4 km (negative because it's in the opposite direction)
3. Next, add up the horizontal and vertical components separately to find the total x and y distances from the starting point.
Total horizontal distance:
x = x₁ + x₂ ≈ 6.062 km + 6.928 km ≈ 13.99 km
Total vertical distance:
y = y₁ + y₂ ≈ 3.5 km - 4 km ≈ -0.5 km
4. Calculate the distance and bearing from the starting point to the final destination using the Pythagorean theorem and trigonometry.
Distance (d) from the starting point:
d = √(x² + y²) ≈ √((13.99 km)² + (-0.5 km)²) ≈ √(195.20 km² + 0.25 km²) ≈ √195.45 km ≈ 13.96 km
Bearing (θ) from the starting point:
θ = atan(y/x) ≈ atan((-0.5 km)/(13.99 km)) ≈ -1.97°
Note: The negative angle indicates a bearing in the opposite direction, so we need to adjust it to a positive angle.
θ = 360° - 1.97° ≈ 358.03°
Therefore, the distance from the girl's starting point is approximately 13.96 km, and the bearing is approximately 358.03°.