A scout troop is practicing its orienteering skills with map and compass. First they walk due east for 1.5 km. Next, they walk 32° west of north for 3.5 km. How far and in what direction must they walk to go directly back to their starting point?
X=hor = 1.5 + 3.5cos(90+32) = -0.355km.
Y = ver = 3.5sin(90+32) = 3.01km.
d = X/cosA = -0.355 / cos96.7 = 3.04km@
96.7deg.=Distance from starting point.
To determine the distance and direction the scout troop must walk to get back to their starting point, we can use vector addition.
First, let's break down their movements into east and north components.
1. The distance they walked due east can be considered as having an east component of 1.5 km and a north component of 0 km, since they didn't move north.
2. The distance they walked 32° west of north can be split into east and north components. To find these components, we can use trigonometry.
The north component can be calculated as: 3.5 km * cos(32°).
The east component can be calculated as: 3.5 km * sin(32°).
Next, we sum up the east and north components:
East component: 1.5 km + 3.5 km * sin(32°)
North component: 0 km + 3.5 km * cos(32°)
To find the total distance they need to walk back to the starting point, we can use the Pythagorean theorem:
Total distance = √(East component^2 + North component^2)
Finally, to determine the direction, we can use the inverse tangent function:
Direction = arctan(North component / East component)
Let's calculate the values:
East component: 1.5 km + 3.5 km * sin(32°) ≈ 3.01 km
North component: 0 km + 3.5 km * cos(32°) ≈ 2.98 km
Total distance = √(3.01 km^2 + 2.98 km^2) ≈ 4.23 km
Direction = arctan(2.98 km / 3.01 km) ≈ 42.4°
Therefore, they need to walk approximately 4.23 km in a direction of about 42.4° to go directly back to their starting point.