Trigonometry problem:

The sides AB and AC of triangle ABC are of lengths 4 and 7 respectively. M is the midpoint of BC. AM is of length 3.5, Find the length of BC.

Complete the parallelogram ABPC , where BP=7 and PC=4

In a ||gram , diagonals bisect each other, so M is the midpoint of both diagonals and AP = 2(3.5) = 7

(looks like we have isosceles triangles, but that doesn't matter)

let angle PAC =Ø
let MC = a , thus BC = 2a for later

In triangle APC, by cosine law
4^2 = 7^2 + 7^2 - 2(7)(7)cosØ
cosØ = 82/98 = 41/49

in triangle AMC
a^2 = 3.5^2 + 7^2 - 2(3.5)(7)cosØ
= 61.25 - 49(41/49) = 20.25
a=√20.25 = 4.5

So BC = 2a = 2(4.5) = 9

better check my arithmetic, I have made some silly errors lately.

Thank You!

To find the length of BC, we can use the concept of the midpoint of a line segment. We can start by drawing a diagram of triangle ABC, where AB = 4, AC = 7, and AM = 3.5. Let's denote the length of BC as x.

According to the midpoint theorem, the length of one side of a triangle, in this case, AM, is equal to half the sum of the lengths of the other two sides. In formula form:

AM = (AB + AC) / 2

Substituting the known values:

3.5 = (4 + 7) / 2

Next, we can solve this equation to find the value of BC, which is x.

Multiply both sides of the equation by 2:

7 = 4 + 7

Now, simplify the equation:

7 = 11

However, we can see that the equation 7 = 11 is not true.

This means that the given information does not form a valid triangle. There might be an error in the problem statement or values provided. Make sure to double-check the problem to ensure that it is correct.