Question

In a triangle \[ABC\], if \[c^{2} + a^{2} - b^{2} = ac\], then find the value of \[\angle B\].
A. \[\dfrac{\pi }{6}\]
B. \[\dfrac{\pi }{4}\]
C. \[\dfrac{\pi }{3}\]
D. None of these

Answer 1

Hint:
First, find the value of angle \[B\] by using the law of cosines for any triangle. Solve the equation by substituting the given equation to get the required answer.



Formula Used:
Laws of cosines:
\[\cos A = \dfrac{{b^{2} + c^{2} - a^{2}}}{{2bc}}\]
\[\cos B = \dfrac{{c^{2} + a^{2} - b^{2}}}{{2ac}}\]
\[\cos C = \dfrac{{a^{2} + b^{2} - c^{2}}}{{2ab}}\]
Value cosine angle:
\[\cos\dfrac{\pi }{3} = \dfrac{1}{2}\]



Complete step-by-step answer:
Given:
In a triangle \[ABC\], \[c^{2} + a^{2} - b^{2} = ac\].

Let’s apply the law of cosine for the angle \[B\].
\[\cos B = \dfrac{{{c^2} + a^{2} - b^{2}}}{{2ac}}\]
Substitute the value of \[{c^2} + a^{2} - b^{2}\] from the given equation.
\[\cos B = \dfrac{{ac}}{{2ac}}\]
\[ \Rightarrow \cos B = \dfrac{1}{2}\]
\[ \Rightarrow \cos B = \cos\dfrac{\pi }{3}\]
Equate the angles.
\[B = \dfrac{\pi }{3}\]
Hence the correct option is C.

Additional information:
Cosine law states that: The square of one side of a triangle is equal to the sum of squares the remaining sides subtracted by twice the product of the sides and cosine of the angle between the sides.



Note:
The law of cosines is used to find an unknown side of a triangle given the value of two sides and their included angle or to find an unknown angle given three sides of a triangle.