# In$\Delta ABC,{\text{ }}2ac\sin \left( {\dfrac{{A - B + C}}{2}} \right)$is equal to:

$

{\text{a}}{\text{. }}{a^2} + {b^2} - {c^2} \\

{\text{b}}{\text{. }}{{\text{c}}^2} + {a^2} - {b^2} \\

{\text{c}}{\text{. }}{{\text{b}}^2} - {a^2} - {c^2} \\

{\text{d}}{\text{. }}{{\text{c}}^2} - {a^2} - {b^2} \\

$

Last updated date: 19th Mar 2023

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Answer

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309.9k+ views

Hint: Sum of all the angles in a triangle is $180^0$

We have to find out the value of

$2ac\sin \left( {\dfrac{{A - B + C}}{2}} \right)......................\left( 1 \right)$

As we know in triangle the sum of all angles is always equal to$180^\circ $

$

\Rightarrow A + B + C = 180^\circ \\

\Rightarrow A + C = 180^\circ - B \\

$

Therefore from equation 1

$2ac\sin \left( {\dfrac{{180^\circ - B - B}}{2}} \right) = 2ac\sin \left( {\dfrac{{180^\circ - 2B}}{2}} \right) = 2ac\sin \left( {90^\circ - B} \right)$

Now, we know that$\sin \left( {90^\circ - B} \right) = \cos B$, so, use this property

$ \Rightarrow 2ac\sin \left( {90^\circ - B} \right) = 2ac\cos B.................\left( 2 \right)$

Now as we know in any triangle the cosine of angle$B$is

i.e.$ \Rightarrow \cos B = \left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right)$

Therefore from equation 2

\[

\Rightarrow 2ac\cos B = 2ac\left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right) \\

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

\]

Hence, option b is correct.

Note: -In such types of questions the key concept we have to remember is that the sum of all angles in any triangle is always$180^\circ $ and also remember the formula of cosine of any angle in a triangle, then simplify we will get the required answer.

We have to find out the value of

$2ac\sin \left( {\dfrac{{A - B + C}}{2}} \right)......................\left( 1 \right)$

As we know in triangle the sum of all angles is always equal to$180^\circ $

$

\Rightarrow A + B + C = 180^\circ \\

\Rightarrow A + C = 180^\circ - B \\

$

Therefore from equation 1

$2ac\sin \left( {\dfrac{{180^\circ - B - B}}{2}} \right) = 2ac\sin \left( {\dfrac{{180^\circ - 2B}}{2}} \right) = 2ac\sin \left( {90^\circ - B} \right)$

Now, we know that$\sin \left( {90^\circ - B} \right) = \cos B$, so, use this property

$ \Rightarrow 2ac\sin \left( {90^\circ - B} \right) = 2ac\cos B.................\left( 2 \right)$

Now as we know in any triangle the cosine of angle$B$is

i.e.$ \Rightarrow \cos B = \left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right)$

Therefore from equation 2

\[

\Rightarrow 2ac\cos B = 2ac\left( {\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}} \right) \\

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

\]

Hence, option b is correct.

Note: -In such types of questions the key concept we have to remember is that the sum of all angles in any triangle is always$180^\circ $ and also remember the formula of cosine of any angle in a triangle, then simplify we will get the required answer.

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