In a \[\Delta ABC, 2a\sin (\dfrac{A-B+C}{2})\] is equal to [IIT Screening 2000]
A. ${{a}^{2}}+{{b}^{2}}-{{c}^{2}}$
B. ${{c}^{2}}+{{a}^{2}}-{{b}^{2}}$
C. ${{b}^{2}}-{{c}^{2}}-{{a}^{2}}$
D. ${{c}^{2}}-{{a}^{2}}-{{b}^{2}}$
Answer
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Hint:
In this case, we will use the angle sum property of a triangle according to which the triangle's triangle's interior angles add up to $180^{o}$. Using this relation and putting it in the given scenario we will get the result.
Formula Used:
Three sides in the triangle $ABC$ are represented by the letters $AB, BC,$ and $CA$. The three vertices are $A, B$, and $C$, and the three interior angles are $\angle ABC, \angle BCA,$ and $\angle CAB$. Then according to the angle sum property of the triangle: $\angle ACB + \angle BAC + \angle CBA= 180^{o}$
Laws of cosine for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$b^2 = a^2 +c^2 − 2ac.\cos B$.
Complete step-by-step solution:
We have, $2a\sin (\dfrac{A-B+C}{2})$
Since $ABC$ is a triangle, therefore we can say;
$A+B+C=180^{o}\\
A+C=180^{o}-B …(i)$
Now consider
$2a\sin (\dfrac{A-B+C}{2})$
Substitute value of $A+C$ using eq (i)
$\Rightarrow2a\sin \,\,\left( \dfrac{180^{o}-B-B}{2} \right)\\
\Rightarrow2a\sin(\dfrac{180^{o}-2B}{2})\\
\Rightarrow 2ac \sin(90^{o}-B)
\Rightarrow 2ac \cos B$
Using the cosine rule of the triangle we get;
$\therefore b^2 = a^2 +c^2 − 2ac.\cos B\\
\therefore \cos B=\dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow 2ac \dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow a^2+c^2-b^2$
$2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$.
Hence, $2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$. So, option B is correct .
Note:
Keep in mind that such type of question requires knowledge of the basic trigonometric conversation. Here we have uses the conversion of sine at $(90^{o}-B)$ which turns out to be $\cos B$ as it lies in the first quadrant so it is taken as positive. Using the law of cosine for angle B, students usually make mistakes in taking the wrong angles. Different angles use cosine rules in different ways.
In this case, we will use the angle sum property of a triangle according to which the triangle's triangle's interior angles add up to $180^{o}$. Using this relation and putting it in the given scenario we will get the result.
Formula Used:
Three sides in the triangle $ABC$ are represented by the letters $AB, BC,$ and $CA$. The three vertices are $A, B$, and $C$, and the three interior angles are $\angle ABC, \angle BCA,$ and $\angle CAB$. Then according to the angle sum property of the triangle: $\angle ACB + \angle BAC + \angle CBA= 180^{o}$
Laws of cosine for triangle $ABC$ whose length is $a, b$, and $c$ respectively is given by;
$b^2 = a^2 +c^2 − 2ac.\cos B$.
Complete step-by-step solution:
We have, $2a\sin (\dfrac{A-B+C}{2})$
Since $ABC$ is a triangle, therefore we can say;
$A+B+C=180^{o}\\
A+C=180^{o}-B …(i)$
Now consider
$2a\sin (\dfrac{A-B+C}{2})$
Substitute value of $A+C$ using eq (i)
$\Rightarrow2a\sin \,\,\left( \dfrac{180^{o}-B-B}{2} \right)\\
\Rightarrow2a\sin(\dfrac{180^{o}-2B}{2})\\
\Rightarrow 2ac \sin(90^{o}-B)
\Rightarrow 2ac \cos B$
Using the cosine rule of the triangle we get;
$\therefore b^2 = a^2 +c^2 − 2ac.\cos B\\
\therefore \cos B=\dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow 2ac \dfrac{a^2+c^2-b^2}{2ac}\\
\Rightarrow a^2+c^2-b^2$
$2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$.
Hence, $2a\sin \,\,\left( \dfrac{A-B+C}{2} \right)=a^2+c^2-b^2$. So, option B is correct .
Note:
Keep in mind that such type of question requires knowledge of the basic trigonometric conversation. Here we have uses the conversion of sine at $(90^{o}-B)$ which turns out to be $\cos B$ as it lies in the first quadrant so it is taken as positive. Using the law of cosine for angle B, students usually make mistakes in taking the wrong angles. Different angles use cosine rules in different ways.
Last updated date: 24th Sep 2023
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