Starting from one oasis, a camel walks 25km in a direction 30 degrees south of west and then walks 30km toward the north to a second oasis. What distance separates the two oases?

I made a quick sketch and found the angle between the two paths to be 60°

So it becomes a simple Cosine Law

x^2 = 25^2 + 30^2 - 2(25)(30)cos60
= 775
x = √775 = 27.84 appr.

50degrees

Well, if the camel is going in a direction 30 degrees south of west, it's safe to say it might need a GPS... or some dance lessons! Anyway, let's crunch the numbers.

The camel walks 25km in this wobbly direction, but since it's heading south of west, we'll have to split that into two components: one toward the west and one toward the south.

To find the westward component, we multiply the distance (25km) by the cosine of 30 degrees. Cosine of 30 degrees is approximately 0.866. So, 25km * 0.866 is about 21.65km to the west.

To find the southward component, we multiply the distance (25km) by the sine of 30 degrees. Sine of 30 degrees is approximately 0.5. So, 25km * 0.5 is about 12.5km to the south.

Now, we need to add this to the northward journey of 30km to reach the second oasis.

So, the distance separating the two oases is about 21.65km + 30km + 12.5km. That gives us a grand total of approximately 64.15km.

So, it seems the camel went on quite the little adventure here! 64.15km is the distance separating the two oases.

To find the distance between the two oases, we can use the concept of vector addition. Let's break down the camel's journey into two components:

1. Walking 25km in a direction 30 degrees south of west: Convert this direction into a vector. To do so, we need to find the horizontal and vertical components of this vector.

The horizontal component is calculated by multiplying the magnitude (25km) by the cosine of the angle (30 degrees south of west). The vertical component is determined by multiplying the magnitude (25km) by the sine of the angle.

Horizontal component = 25km * cos(30 degrees) = 25km * √3/2 ≈ 21.65km
Vertical component = 25km * sin(30 degrees) = 25km * 1/2 = 12.5km

So the first leg of the journey can be represented by the vector [-21.65km, -12.5km].

2. Walking 30km toward the north: This leg can be represented by the vector [0km, 30km].

Now, to find the displacement between the two oases, we add the two vectors together:

Displacement = [-21.65km, -12.5km] + [0km, 30km]

Adding the horizontal and vertical components separately:

Horizontal component = -21.65km + 0km = -21.65km
Vertical component = -12.5km + 30km = 17.5km

The displacement vector is then [-21.65km, 17.5km].

Finally, we can use the Pythagorean theorem to calculate the distance between the two oases. The distance is the magnitude of the displacement vector:

Distance = √((-21.65km)^2 + (17.5km)^2)
= √(468.7225km^2 + 306.25km^2)
= √774.9725km^2
≈ 27.85km

Therefore, the distance between the two oases is approximately 27.85 kilometers.

Take x as east, and y as north, and the starting position as (0,0).

The coordinates of the first stop are
(-25cos(30°), -25sin(30°)
The second stop is simply 30km towards +y, or
(-25cos(30°), 30-25sin(30°)
=(21.65,17.5)

Use Pythagoras theorem to calculate the distance from the original point (0,0).