Mark the correct alternative in each of the following:

In any ∆ABC, 2(bc cosA + ca cosB + ab cosC) =

(a) $a b c$ (b) $a + b + c$ (c) $a^{2} + b^{2} + c^{2}$ (d) $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}$

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Solution

Using cosine rule, we have

$2 (b c \cos A + c a \cos B + a b \cos C) = 2 b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) + 2 c a (\frac{c^{2} + a^{2} - b^{2}}{2 c a}) + 2 a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) = b^{2} + c^{2} - a^{2} + c^{2} + a^{2} - b^{2} + a^{2} + b^{2} - c^{2} = a^{2} + b^{2} + c^{2}$

Hence, the correct answer is option (c).

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Similar questions

2(bc cos A+ ca cos B + ab cos C)=

Mark the correct alternative in each of the following :

$I n a n y Δ A B C, 2 (b c c o s A + c a c o s B + a b c o s C) =$

a (b \cos C - c \cos B) =

a^{2}

b^{2} - c^{2}

b^{2} + c^{2}

A B C

2 (b c cos A + c a cos B + a b cos C) = a^{2} + b^{2} + c^{2}

$2 (b c c o s A + c a c o s B + a b c o s C) = a^{2} + b^{2} + c^{2}$

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