MCQ
In any $\triangle\text{ABC},\text{a}(\text{b}\cos\text{C}-\text{c}\cos\text{B})=$
  • A
    a2
  • B
    b- c2
  • C
    0
  • D
    b+ c2

Answer

  1. b- c2

Solution:

Using cosine rule, we have

$\text{a(b}\cos\text{C}-\text{c}\cos\text{B})$

$=\text{ab}\Big(\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}\Big)-\text{ca}\Big(\frac{\text{c}^2+\text{a}^2-\text{b}^2}{2\text{ca}}\Big)$

$=\frac{\text{a}^2+\text{b}^2-\text{c}^2-\text{c}^2-\text{a}^2+\text{b}^2}{2}$

$=\frac{2\text{b}^2-2\text{c}^2}{2}$

$=\text{b}^2-\text{c}^2$

Hence, the correct answer is option (b).

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