In the given triangle ABC, what is the length of BC? (The given triangle is not rt. angled triangle)
Let the length of the third side (i.e. BC) of the triangle be a.
Applying cosine rule for triangles, `cosA = (b^2+c^2-a^2)/(2bc)` ,
`rArra^2 = b^2+c^2-2b*c*cosA` ,
Putting the values from the given triangle, we get
`a^2 = 7^2+(4sqrt2)^2-2*7*4sqrt2*cos45°`
Therefore `a = sqrt25 = 5` units.
The length of the third side (i.e. BC) of the given triangle is 5 units.