In any ,2(bc + ca + ab )= 1. abc 2. a + b + c 3. a^2+b^2+c^2 4. 1 a^2 +1 b^2 +1 c^2

3. a^2+b^2+c^2 Solution: Using cosine rule, we have 2(bc +ca +ab ) =2bc ( b^2+c^2-a^2 2bc )+2ca ( c^2+a^2-b^2 2ca )+2ab ( a^2+b^2-c^2 2ab ) =b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2 =a^2+b^2+c^2 Hence, the correct answer is option (c).

In any ,2(bc + ca + ab )= 1. abc 2. a + b + c 3. a^2+b^2+c^2 4. 1…

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Question

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

$\text{abc}$
$\text{a + b + c}$
$\text{a}^2+\text{b}^2+\text{c}^2$
$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}+\frac{1}{\text{c}^2}$

✓

Answer

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

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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