A BOAT SAILLING ON A BEARING OF 070 FROM A POINT Y,8KM AWAY FROM X.
THEREFORE,ON A BEARING OF 145, IT GETS TO A POINT Z, 15KM AWAY Y. WHAT IS ITS DISTANCE AND BEARING FROM X WHILE AT POINT Z
To find the distance and bearing of the boat at point Z from point X, we can use the concept of vectors.
First, let's consider the position of the boat at point Y. We can represent this as a vector, let's call it VY, with a magnitude of 8 km and a direction of 070°.
Next, let's find the position of the boat at point Z relative to point Y. We can represent this as another vector, let's call it VZY. The magnitude of VZY is 15 km, and its direction is 145°.
To find the position of the boat at point Z relative to point X, we need to add these two vectors together. This can be done using vector addition.
Let VXY be the vector representing the position of the boat at point Z relative to point X.
VXY = VY + VZY
Using vector addition, we can break down the vectors into their horizontal and vertical components:
VY_horizontal = VY_magnitude * cos(VY_direction)
VY_vertical = VY_magnitude * sin(VY_direction)
VZY_horizontal = VZY_magnitude * cos(VZY_direction)
VZY_vertical = VZY_magnitude * sin(VZY_direction)
Once we have the horizontal and vertical components of VY and VZY, we can add them together to get the horizontal and vertical components of VXY.
VXY_horizontal = VY_horizontal + VZY_horizontal
VXY_vertical = VY_vertical + VZY_vertical
The magnitude of VXY can be calculated using the Pythagorean theorem:
VXY_magnitude = √(VXY_horizontal^2 + VXY_vertical^2)
The bearing of VXY can be found using inverse tangent:
VXY_direction = arctan(VXY_horizontal / VXY_vertical)
Now, let's calculate the values:
VY_horizontal = 8 km * cos(70°) = 8 km * 0.3420 ≈ 2.736 km
VY_vertical = 8 km * sin(70°) = 8 km * 0.9397 ≈ 7.518 km
VZY_horizontal = 15 km * cos(145°) = 15 km * -0.5736 ≈ -8.604 km
VZY_vertical = 15 km * sin(145°) = 15 km * 0.8192 ≈ 12.288 km
VXY_horizontal = 2.736 km + (-8.604 km) ≈ -5.868 km
VXY_vertical = 7.518 km + 12.288 km ≈ 19.806 km
VXY_magnitude = √((-5.868 km)^2 + (19.806 km)^2) ≈ 20.725 km
VXY_direction = arctan((-5.868 km) / (19.806 km)) ≈ -16.944° (assuming angle is measured clockwise from the positive y-axis)
Therefore, the distance of the boat from point X at point Z is approximately 20.725 km, and the bearing (measured clockwise from the positive y-axis) is approximately -16.944°.