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Properties of Triangles

Section: Mathematics

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Question : 6 of 50

Marks: +1, -0

In

\triangle ABC

a: b: c=4: 5: 6

\;\frac{\cos A+3\cos C}{\cos B}=

[AP EAPCET 23 May 2025 S1]

1
4
2
3

Solution:

Given,

a: b: c=4: 5: 6

We can express the sides as

a=4x, b=5x

and

c=6x

for some positive constant

x

.

\;\because \cos A=\;\frac{b^2+c^2-a^2}{2bc}(\;\text{ By Cosine rule }\;)

\;=\;\frac{(5x)^2+(6x)^2-(4x)^2}{2 \times 5x \times 6x}=\;\frac{45x^2}{60x^2}=\;\frac{3}{4}

Similarly,

\cos B=\;\frac{(4x)^2+(6x)^2-(5x)^2}{2(4x)(6x)}

=\;\frac{27x^2}{48x^2}=\;\frac{9}{16}

and

\cos C=\;\frac{(4x)^2+(5x)^2-(6x)^2}{2(4x)(5x)}

=\;\frac{5x^2}{40x^2}=\;\frac{1}{8}

So,

\;\frac{\cos A+3\cos C}{\cos B}=\;\frac{\;\frac{3}{4}+3\left(\frac{1}{8}\right)}{\frac{9}{16}}=\;\frac{\;\frac{6+3}{8}}{\frac{9}{16}}

=\;\frac{9}{8} \times \;\frac{16}{9}=2

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