The treasure map in the figure gives the following directions to the buried trea

Question

The treasure map in the figure gives the following directions to the buried treasure: "Start at the old oak tree, walk due north for 530 paces, then due east for 110 paces. Dig." But when you arrive, you find an angry dragon just north of the tree. To avoid the dragon, you set off along the yellow brick road at an angle 60? east of north. After walking 400 paces you see an opening through the woods.

Which direction should you go to reach the treasure?

How far should you go to reach the treasure?

Explanation / Answer

Let A = 530 paces due north

We write in vector form (with north as +y, east as +x)

A = 530 j

Let B = 110 due east

B = 110 i

Vector sum, A + B represents the net displacement to the point of treasure.

Now instead we walk 400 paces at the direction 60 degrees east of north.

Let this displacement be represented by vector C

The angle between (North) y axis and C (yellow road) is 60 degrees.

Thus angle between x axis and C = 90 - 60 = 30 degrees.

The vector C can be written as:

C = 400 cos 30 i + 400 sin 30 j = 346.4 i + 200 j

Let let D represent the remaing displacement we have to walk to get to treasure.

Thus, C + D must reach us to treasure

or,

C+D = A+B

or,

D = A+B-C

This is vector operations, so we write as:

D = (0 i + 530j) + (110 i + 0 j) - (346.4 i + 200 j)

D = -236.4 i + 330 j

This is the displacement we have to undertake to reach the treasure.

Now we calculate its magnitude (how far to go) and angle (what direction to go):

How far to go:

Magnitude of D :

square root ((-236.4)2 + 3302) = square root (55884.96+108900) = 405.94 paces = 406 paces you need to walk more.

Which direction to go:

tan (theta) = y/x = -236.4/330 = -0.72

or,

theta = -35.8 degrees = 35.8 degrees with respect to -x axis

or, you have to walk in North-West direction, or, 35.8 degrees north of west = 90-35.8 = 54.2 degrees west of north to reach the treasure.

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