In triangle ABC, prove the following: \[a^2 \sin \left( B - C \right) = \left( b^2 - c^2 \right) \sin A\]

Let \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\] Consider the RHS of the equation \[a^2 \sin \left( B - C \right) = \left( b^2 - c^2 \right) \sin A\] \[RHS = k^2 \sin A\left( \sin^2 B - \sin^2 C \right) \]\[ = k^2 \sin A\left[ \sin\left( B + C \right)\sin\left( B - C \right) \right] \left[ \because \sin^2 B - \sin^2 C = \sin\left( B + C \right)\sin\left( B - C \right) \right]\]\[ = k^2 \sin A\left[ \sin\left( \pi - A \right)\sin\left( B - C \right) \right] \left[ \because A + B + C = \pi \right]\]\[ = k^2 \sin A\left[ \sin\left( A \right)\sin\left( B - C \right) \right]\]\[ = k^2 \sin^2 A\sin\left( B - C \right)\]\[ = a^2 \sin\left( B - C \right) = LHS \left[ \because a = k\sin A \right]\]\[\text{ Hence proved } .\]

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

प्रश्न

In triangle ABC, prove the following:

a^{2} \sin (B - C) = (b^{2} - c^{2}) \sin A

उत्तर

Let

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

Consider the RHS of the equation

a^{2} \sin (B - C) = (b^{2} - c^{2}) \sin A

$R H S = k^{2} \sin A (\sin^{2} B - \sin^{2} C)$
$= k^{2} \sin A [\sin (B + C) \sin (B - C)] [∵ \sin^{2} B - \sin^{2} C = \sin (B + C) \sin (B - C)]$
$= k^{2} \sin A [\sin (π - A) \sin (B - C)] [∵ A + B + C = π]$
$= k^{2} \sin A [\sin (A) \sin (B - C)]$
$= k^{2} \sin^{2} A \sin (B - C)$
$= a^{2} \sin (B - C) = L H S [∵ a = k \sin A]$
$Hence proved .$

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Sine and Cosine Formulae and Their Applications

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 12 Q 11 Q 13

अध्याय 10: Sine and cosine formulae and their applications - Exercise 10.1 [पृष्ठ १३]

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आरडी शर्मा Mathematics [English] Class 11

अध्याय 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 12 | पृष्ठ १३

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

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