Question

The expedition team plan a final practice run to test the range of their communication equipment. One member travels a distance of 12km due north. Another team member heads 50degree east of north and travels a distance of 10km. How far apart are the two team members?

Answer 1

The two team members are approximately 16.64km apart.

Step-by-step explanation:

To find the distance between the two team members, we can represent their displacements as vector components. One member travels 12km due north, so their displacement is 12km in the +y direction. The other member travels 10km at an angle of 50 degrees east of north. We can break down this displacement into its x and y components using trigonometry.

The x-component can be found as 10km * cos(50 degrees) and the y-component can be found as 10km * sin(50 degrees). The x-component is approximately 6.45km and the y-component is approximately 7.66km. Now, we can calculate the distance between the two team members using the Pythagorean theorem.

Distance = sqrt((12km + 6.45km)^2 + (7.66km)^2) = sqrt(218.2km^2 + 58.79km^2) = sqrt(276.99km^2) = 16.64km.

Therefore, the two team members are approximately 16.64km apart.