Question

In any △ ABC , prove that
$\rm \: a(b \cos C \: - c \cos B) = {b}^{2} - {c}^{2}$
​

Answer 1

$\bigstar \ \underline{\underline{ \bf{Question}}}$

In any △ ABC , prove that

$\sf \: a(b \cos C \: - c \cos B) = {b}^{2} - {c}^{2}$

Solution

Using the cosine formula , we have

Since we have to do proving so Let's take LHS first and make it equal to the right hand side .

Taking L.H.S

⇒ a ( b cosC - C cosB )

$⇒\footnotesize\sf \: ab. \dfrac{ {a}^{2} + {b}^{2} - {c}^{2}}{2ab} - ac. \dfrac{ {c}^{2} + {a}^{2} - {b}^{2}}{2ca}$

$⇒ \footnotesize \sf\dfrac{1}{2}( {a}^{2} + {b}^{2} - {c}^{2} - {c}^{2} - {a}^{2} + {b}^{2})$

$⇒ \rm\dfrac{1}{2} (2 {b}^{2} - 2 {c}^{2} ) = \boxed{ \tt\pink{{b}^{2} - {c}^{2}} }$

Therefore, LHS = b² - c² & RHS = b² - c²

$\mid \sf{LHS = RHS } \mid$

$\sf\: Hence \: , \: Proved ✔$

$\rule{200pts}{2pts}$

$\odot \: \underline{ \sf{Related \: Information }}$

❶ The Laws of Sines ( Sine Law ) :-

In any triangle, the sides are proportional to the sines of the opposite angles i.e

$\twoheadrightarrow \sf \dfrac{ {a}^{2} }{ \sin \sf \: A} = \dfrac{ {b}^{2} }{ \sin \sf \: B} = \dfrac{ {c}^{2} }{ \sin \sf \: C}$

❷ Projection Law :-

$\twoheadrightarrow\boxed{\begin{array}{c}\sf \: a = b \cos C + c \cos B \\ \sf \:b = c \cos A + a \cos C \\ \sf \:c = a \cos B + b \cos A\end{array}}$

❸ The Laws of Cosines :-

The square on any side of a triangle is equal to the sum of the squares on the other two sides minus twice the product of those two sides and the cosine of the included angle , i.e.

$\twoheadrightarrow\boxed{ \begin{array} { |c|} \sf \: a² = b² + c² - 2bc cosA \\ \sf \: b² = c² + a² - 2ca cosB \\ \sf c² = a² + b² - 2ab cosC \end{array}}$

❹ Area of Triangle :-

The area of triangle is given by

$\sf \triangle = \dfrac{1}{2}ab \sin C = \dfrac{1}{2} bc \sin A = \dfrac{1}{2}ac \sin B$

$\rule{200pts}{2pts}$

Thankyou

Answer 2

Question

$\sf \: In \: any \: △ ABC , \: prove \: that \\ \rm \: a(b \cos C \: - c \cos B) = {b}^{2} - {c}^{2}$

Solution

a( b cosC - c cosB )

ab. a² + b² - c² / 2ab - ac . c² + a² - b² / 2ca

= 1/2 ( a² + b² - c² - c² - a² + b² )

= 1/2 ( 2b² - 2c² )

= b² - c²

LHS = RHS

Hence Proved

Thankyou