General

yea my bad. i shouldn't do math problems at 3 a.m.

[quote=dr4g0ns1ay3r]I just use the 30 60 90 formula. 70 is the Hypotenuse. 50 is one of the side.

x^2+50^2=70^2------x^2+2500=4900------x^2=2400----x=about 49m or 48.9898 90 degree to the east

But the problem i fear is that the answer is suppose to be a whole number not rounded... Wait for someone else that has a different solution[/quote]

You assumed that the child follows a right triangular path which is something that we cannot do.

The method I used was to relate the boy's path to a Cartesian plane such that north was in the positive y-direction and east was in the positive x-direction. I then solve for the horizontal and vertical components of the child's initial displacement vector. These components are given by the equations:
x-direction: x(initial) = 50cos210
y-direction: y(initla) = 50sin210

Since the child has a final displacement vector that points due south, we can assume that the net distance traveled in the horizontal direction will be equal to zero; thus, his total distance traveled in the x-direction for the first portion of his walk will be equal in magnitude but opposite in direction to the total distance traveled in the x-direction for the second part of his walk.
lx(inital)l = lx(final)l
l50cos210l = l-50cos210l = -25sqrt3
The boy must walk 25sqrt3 m in the horizontal direction in order to be able to arrive at any position directly south of his origin.

Now that we have solved the horizontal axis, we must now solve for the distance that the child must travel in order to arrive at a position 70 m south of his origin (represented as -70 on the Cartesian plane).

The initial distance traveled in the vertical direction is represented by the equation y(initial) = 50sin210 = -25 m
The child travels 25 m in the southern direction. Since our goal is 70 m due south, simple math will show that the child will need to travel 45 m in the negative vertical direction in order to reach his intended destination.

To recap, the child needs to travel a total of 25sqrt3 m directly east and 45 m directly south before reaching his destination of 70 m due south relative to his origin.
You can use Pythagorean theorem and the tan function to find the direction the boy needs to walk in and the magnitude of his second displacement vector.

The direction of travel will be given by: tan^-1 theta = (-45/25sqrt3) = -46.10 degrees.

The magnitude of the boy's second displacement vector(v) is given by: lvl = sqrt[45^2 + (25sqrt3)^2] = 10sqrt39 = 62.45 m.

The child needs to walk 62.45 m 46.10 degrees south of east in order to reach his destination.

PS: my apologies if this is extremely confusing; I wrote this very at a very late time in the evening.

[quote=dr4g0ns1ay3r] I just use the 30 60 90 formula. 70 is the Hypotenuse. 50 is one of the side.

x^2+50^2=70^2------x^2+2500=4900------x^2=2400----x=about 49m or 48.9898 90 degree to the east

But the problem i fear is that the answer is suppose to be a whole number not rounded... Wait for someone else that has a different solution [/quote]

I think it would be better if you use laws of cosine to figure out the distance and then use law of sine to find the angle because how do you know the other two angles are 60 and 90. The problem only gives you a 30 degrees angle.

I just use the 30 60 90 formula. 70 is the Hypotenuse. 50 is one of the side.

x^2+50^2=70^2------x^2+2500=4900------x^2=2400----x=about 49m or 48.9898 90 degree to the east

But the problem i fear is that the answer is suppose to be a whole number not rounded... Wait for someone else that has a different solution

How would you solve it

this is why I hate word problems... its 100% better for me to see the diagram >.<

Please don't paraphrase the problem. Every detail is important...

edit: this is more like a geometry problem.