In the given triangle ABC, what is the length of BC? (The given triangle is not rt. angled triangle)
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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.
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