A ship sails 95 km on a bearing of 140 degree,then a further 102 km on a bearing of 260 degree and then returns directly to its starting point.find the length and bearing of the return journey.
I labeled my triangle ABC, with A as the starting point
AB = 95, and BC = 102 , and angle ABC = 60°
By the cosine law:
AC² = 95² + 102² - 2(95)(102)cos60
= 98.686
= 99 km
why 98^2
To find the length and bearing of the return journey, we need to determine the coordinates of the starting point and the ending point of each leg of the journey.
First, let's calculate the coordinates of the starting point (S) and the ending point (E1) of the first leg:
- Starting point (S): We assume the starting point as the origin, with coordinates (0, 0).
- Ending point of the first leg (E1): To find the coordinates of E1, we use trigonometry. Since the ship sails 95 km on a bearing of 140 degrees, we can break down the movement into horizontal and vertical components. The horizontal component can be calculated as 95 * cos(140) and the vertical component as 95 * sin(140). Adding these components to the starting point's coordinates gives us the coordinates of E1.
Next, let's calculate the coordinates of the ending point (E2) of the second leg:
- Ending point of the second leg (E2): Similar to the first leg, we can calculate the horizontal and vertical components of the second leg using the given distance of 102 km and the bearing of 260 degrees. The coordinates of E2 can be obtained by adding these components to the coordinates of E1.
Finally, we can find the length and bearing of the return journey:
- Length of the return journey: The distance between the starting point (S) and the ending point (E2) will give us the length of the return journey. This can be calculated using the distance formula: √((x2 - x1)^2 + (y2 - y1)^2), where (x1, y1) are the coordinates of S and (x2, y2) are the coordinates of E2.
- Bearing of the return journey: We can find the bearing of the return journey by calculating the angle between the positive x-axis and the line segment connecting S and E2. This can be done using trigonometry and the equation: tan^(-1)((y2 - y1)/(x2 - x1)).
By following these steps, you can calculate the length and bearing of the return journey.
I labeled my triangle ABC, with A as the starting point
AB = 95, and BC = 102 , and angle ABC = 60°
By the cosine law:
AC^2 = 98^2 + 102^2 - 2(95)(102)cos60°
find AC
Once you have AC, use the sine law to find angle BAC and
convert that into a bearing