trigo quest?

Question

ABC is a triangle. AB=18, AC=16, BM=MC=13, AM=?

opppss! forgot to write M is on the line BC.

and you are not allowed a scientific calculator

Answer 1

AM could be anything, so long as it doesn't violate the triangle inequality. Refer to:
http://math.colgate.edu/~kellen/interspace/AM.gif


For ABM, it must be true that:

AB - BM < AM < AB + BM

5 < AM < 31

For triangle ACM, it must be true that:

AC - CM < AM < AC + CM
3 < AM < 29

Combining those two equalities (they BOTH must hold, not an either-or thing), we know:

5 < AM < 29

any such value works for AM.

EDIT:

Based on the additional information, we know a whole lot more about this problem! See this revised diagram:
http://math.colgate.edu/~kellen/interspace/ACM.gif

Based on this information, we can take two approaches. The first would be to modify the solution above and try to work it out from there. That may give us some information, but not all of it.

You see, the triangle inequality isn't going to be the best approach here. There are, of course, no longer multiple solutions. What we're looking for now is just the length of a median. That's not so bad, the formula is simple and known (it follows from the Pythagorean theorem):

AM² = (1/4) (2AB² + 2AC² - BC²)

AM² = (1/4) (2 × 18² + 2 × 16² - 26²)

AM² = (1/4) (484)

AM² = 121

AM = 11

Answer 2

AM = 11

I would drew this. Draw line AB at 18 units. At point A use a compass and draw a partial circle Radius 16. The Point B will be on one of those points on that arc.

Answer 3

Use the cosine rule to find one of the angles of the triangle, then use it again on one of the smaller triangles to find AM.

If the side AB = c, BC = a, and CA = b,

the cosine rule is simply

b^2 = a^2 + c^2 - 2*a*c*cosine(angle B).

I am sure that you are capable of working this out by yourself.

Answer 4

we need one more information...