MCQ
In any $\triangle\text{ABC},2(\text{bc}\cos\text{A + ca}\cos\text{B + ab}\cos\text{C})=$
- A$\text{abc}$
- B$\text{a + b + c}$
- C$\text{a}^2+\text{b}^2+\text{c}^2$
- D$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}$
Solution:
Using cosine rule, we have
$2(\text{bc}\cos\text{A}+\text{ca}\cos\text{B}+\text{ab}\cos\text{C})$
$=2\text{bc}\Big(\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}\Big)+2\text{ca}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)+2\text{ab}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)$
$=\text{b}^2+\text{c}^2-\text{a}^2+\text{c}^2+\text{a}^2-\text{b}^2+\text{a}^2+\text{b}^2-\text{c}^2$
$=\text{a}^2+\text{b}^2+\text{c}^2$
Hence, the correct answer is option (c).
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