In a Δ A B C , a ( b c o s C − c c o s B ) is equal to - Options: A . &nbsp b + c B . &nbsp b - c C . &nbsp b 2 − c 2 D . &nbsp b 2 + c 2

Answer: Option C : C We know from cosine rule c o s C = a 2 + b 2 − C 2 2 a b c o s B = a 2 + c 2 − b 2 2 a c So we'll use these in the given expression. a { b ( a 2 + b 2 − c 2 2 a b ) − c ( a 2 + c 2 − b 2 2 a c ) } a 2 + b 2 − c 2 − a 2 − c 2 + b 2 2 = 2 ( b 2 − c 2 ) 2 = b 2 − c 2

In a ΔABC,a(b cosC −c cos B) is equal to -

Discussion Forum : Solution Of Triangles

Question -

In a $Δ A B C, a (b c o s C - c c o s B)$ is equal to -

Options:

A . b + c

B . b - c

C .

b^{2} - c^{2}

D .

b^{2} + c^{2}

Answer: Option C
:
C
We know from cosine rule

c o s C = \frac{a^{2} + b^{2} - C^{2}}{2 a b}

c o s B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}

So we'll use these in the given expression.

a {b (\frac{a^{2} + b^{2} - c^{2}}{2 a b}) - c (\frac{a^{2} + c^{2} - b^{2}}{2 a c})}

\frac{a^{2} + b^{2} - c^{2} - a^{2} - c^{2} + b^{2}}{2}

= \frac{2 (b^{2} - c^{2})}{2}

= b^{2} - c^{2}

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Discussion Forum : Solution Of Triangles

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