answer the question.?

Question

Two beetles run across flat sand, starting at the same point. Beetle 1 runs 0.52 m due east, then 0.82 m at 28o north of due east. Beetle 2 also makes two runs; the first is 1.71 m at 36o east of due north. What must be the magnitude of its second run if it is to end up at the new location of beetle 1?

Answer 1

Let A be the starting point for the beetles, B be the turning point for the first beetle, C be the finishing point, D be the turning point for the second beetle.

From triangle ABD:
BD^2 = 1.71^2 + 0.52^2 - 2*1.71*0.52cos(36)
BD = 1.325.

sin(ABD) / 1.71 = sin(36) / BD
ABD = 130.66deg.

CBD = 180 - (ABD + 28)
= 21.34deg.

From triangle CBD:
CD^2 = BD^2 + 0.82^2 - 2*BD*0.82*cos(CBD)
CD = 0.64m.

Answer 2

Hi,
This looks like a vector problem, so I'll assume you're familiar with vectors. If you are not, just consider the "i" component and being due east and the j component as being dur north.
Let A be the .52 length vector, B the .82 length, and C the 1.71 length.
Then in rectangular coordinates we have this:
A = .51i + 0j
B = (.82 Cos 28) i + (.82 sin 28)j
...=.724i + .385j
The we have:
A+B = (.52i + 0j) +(.724i +.385j)
=1.24i = .385j

Then:
C = (1.71cos(90-36)) + (1.71sin(90-36))
= 1.0i + 1.384j

Now, to find the difference in rectangular form:
resulatant = AB -AC
=(1.24i +.385j) -(1.0i +1.384j)
=.24i -1.0j

Now, we take the absolute value using the Pythagorean theorem:
Sqrt(.24² +1.01²)
=1.07

Hope this helps.
FE