If $$a, b,c$$ are 3 side of $$△ABC中$$ such that $$(b+c):(c+a):(a+b) = 4:5:6$$, then the maximum internal angle of $$△ABC$$ is

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Uniteasy Admin · Accepted Answer

Let 
$$ b+c=4k$$n
$$c+a=5k$$n
$$a+b=6k$$n
Addition of the three equations
$$2(a+b+c)=15k$$, that is,
$$a+b+c=7.5k$$n
Substituting (1),(2), (3) to (4) gives
$$a=3.5k,b=2.5k,c=1.5k$$
Therefore, the maximum internal angle is A,
Using the law of cosine：
$$\cos A = \dfrac{b^2+c^2-a^2}{2bc}  $$
$$=\dfrac{6.25k^2+2.25k^2-12.25k^2}{7.5k^2}  $$
$$=-1$$
$$\because   A∈(0, 180°)$$
 $$	herefore A=120°$$．
Therefore, the maximum internal angle of $$△ABC$$ is $$120°$$
Option C is the right choice.

Multiple Choice Question (MCQ)

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