Question

In ΔABC, if \dfrac{b+c }{12}=\dfrac{c+a}{13}=\dfrac{a+b}{15} , prove that

\dfrac{\cos A}{2}=\dfrac{\cos B}{7}=\dfrac{\cos C}{11}

Collected in the board: Trigonometry

Steven Zheng posted 3 weeks ago


Answer

Let k be the value of the given ratios, that is

\dfrac{b+c}{12}=\dfrac{c+a}{13}=\dfrac{a+b}{15}=k

Then we get the following equations

b+c=12k

c+a=13k

a+b=15k

Solve the equation system

a=8k

b=7k

c=5k

Using the Law of Cosine

\cos A = \dfrac{b^2+c^2-a^2}{2bc}= \dfrac{1}{7}

\cos B = \dfrac{a^2+c^2-b^2}{2ac}=\dfrac{1}{2}

\cos C = \dfrac{a^2+b^2-c^2}{2ab}=\dfrac{11}{14}

Now,

\cos A:\cos B:\cos C=\dfrac{1}{7}:\dfrac{1}{2}:\dfrac{11}{14} =\dfrac{2}{14}:\dfrac{7}{14}:\dfrac{11}{14}

Therefore,

\dfrac{\cos A}{2}=\dfrac{\cos B}{7}=\dfrac{\cos C}{11}

Steven Zheng posted 3 weeks ago

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