In any ,2(bc + ca + ab )= - Vidyadip

MCQ

In any $\triangle\text{ABC},2(\text{bc}\cos\text{A + ca}\cos\text{B + ab}\cos\text{C})=$

A
$\text{abc}$
B
$\text{a + b + c}$
✓
$\text{a}^2+\text{b}^2+\text{c}^2$
D
$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}$

✓

Answer

Correct option: C.

$\text{a}^2+\text{b}^2+\text{c}^2$

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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