# 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°`

= `49+32-(56sqrt2)/sqrt2`

= `25`

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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