MHT CET · Maths · Properties of Triangles
With usual notations, in \(\triangle \mathrm{ABC}\), if \(\mathrm{a}=2, \mathrm{~b}=3, \mathrm{c}=5\) and \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}\),
then \(\mathrm{k}=\)
- A 6
- B 16
- C 17
- D 12
Answer & Solution
Correct Answer
(D) 12
Step-by-step Solution
Detailed explanation
\(a=2, b=3, c=5\) and \(\frac{\cos A}{a}+\frac{\cos B}{b}+\frac{\cos C}{c}=\frac{k+7}{30}\)
\(\frac{b^{2}+c^{2}-a^{2}}{2 a b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a b c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b c}=\frac{k+7}{30}\)
\(\frac{b^{2}+c^{2}-a^{2}+a^{2}+c^{2}-b^{2}+a^{2}+b^{2}-c^{2}}{2 a b c}=\frac{k+7}{30}\)
\(\begin{aligned} \frac{a^{2}+b^{2}+c^{2}}{2 a b c} &=\frac{k+7}{30} \Rightarrow \frac{2^{2}+3^{2}+5^{2}}{2 \times 2 \times 3 \times 5}=\frac{k+7}{30} \\ \frac{38 \times 30}{60} &=k+7 \Rightarrow 19-7=k \\ k &=12 \end{aligned}\)
\(\frac{b^{2}+c^{2}-a^{2}}{2 a b c}+\frac{a^{2}+c^{2}-b^{2}}{2 a b c}+\frac{a^{2}+b^{2}-c^{2}}{2 a b c}=\frac{k+7}{30}\)
\(\frac{b^{2}+c^{2}-a^{2}+a^{2}+c^{2}-b^{2}+a^{2}+b^{2}-c^{2}}{2 a b c}=\frac{k+7}{30}\)
\(\begin{aligned} \frac{a^{2}+b^{2}+c^{2}}{2 a b c} &=\frac{k+7}{30} \Rightarrow \frac{2^{2}+3^{2}+5^{2}}{2 \times 2 \times 3 \times 5}=\frac{k+7}{30} \\ \frac{38 \times 30}{60} &=k+7 \Rightarrow 19-7=k \\ k &=12 \end{aligned}\)
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