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In Triangle Abc, Prove the Following: a 2 Sin ( B − C ) Sin a + B 2 Sin ( C − a ) Sin B + C 2 Sin ( a − B ) Sin C = 0 - Mathematics

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प्रश्न

In triangle ABC, prove the following: 

\[\frac{a^2 \sin \left( B - C \right)}{\sin A} + \frac{b^2 \sin \left( C - A \right)}{\sin B} + \frac{c^2 \sin \left( A - B \right)}{\sin C} = 0\]

 

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उत्तर

Let 

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] 

Then,
Consider the LHS of the equation

\[\frac{a^2 \sin \left( B - C \right)}{\sin A} + \frac{b^2 \sin \left( C - A \right)}{\sin B} + \frac{c^2 \sin \left( A - B \right)}{\sin C} = 0\]

\[LHS = \frac{a^2 \sin\left( B - C \right)}{\sin A} + \frac{b^2 \sin\left( C - A \right)}{\sin B} + \frac{c^2 \sin\left( A - B \right)}{\sin C}\]
\[ = \frac{k^2 \sin^2 A\sin\left( B - C \right)}{\sin A} + \frac{k^2 \sin^2 B\sin\left( C - A \right)}{\sin B} + \frac{k^2 \sin^2 C\sin\left( A - B \right)}{\sin C} \]
\[ = k^2 \sin A\sin\left( B - C \right) + k^2 \sin B\sin\left( C - A \right) + k^2 \sin C\sin\left( A - B \right) \]
\[ = k^2 \left[ \sin A\left( \sin B\cos C - \sin C\cos B \right) + \sin B\left( \sin C\cos A - \sin A\cos C \right) + \sin C\left( \sin A\cos B - \sin B\cos A \right) \right] \]
\[ = k^2 \left( \sin A\sin B\cos C - \sin A\sin C\cos B + \sin B\sin C\cos A - \sin A\sin B\cos C + \sin A\sin C\cos B - \sin C\sin B\cos A \right)\]
\[ = 0 = RHS\]
\[\text{ Hence proved } .\]

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Sine and Cosine Formulae and Their Applications
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 10: Sine and cosine formulae and their applications - Exercise 10.1 [पृष्ठ १३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 15 | पृष्ठ १३

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