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Given that \[{e^{iA}}\] , \[{e^{iB}}\] , \[{e^{iC}}\] are in A.P. where \[A\] , \[B\], \[C\] are the angles of the triangle , then the triangle is
A.Isosceles
B.Equilateral
C.Right - angled
D.None of these.

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Last updated date: 23rd May 2024
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Answer
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Hint: In this problem, we have to show the type of triangle. First, we need to know the fact that the sum of all the angles of the triangle is \[180^\circ \]. Then, we use the properties of A.P. (Arithmetic progression ) to find the sum of angles of the triangle and also, we can use this trigonometric identity to solve the problem.
 \[\sin A + \sin B = 2\sin \dfrac{{(A + B)}}{2}\cos \dfrac{{(A - B)}}{2}\]
\[\cos A + \cos B = 2\cos \dfrac{{(A + B)}}{2}\cos \dfrac{{(A - B)}}{2}\]
From the degree table, \[\cos 60^\circ = \dfrac{1}{2}\]. Finally, we can get the required solution.

Complete step-by-step answer:
We are given the term \[{e^{iA}}\] , \[{e^{iB}}\] , \[{e^{iC}}\] are in A.P. where \[A\] , \[B\], \[C\] are the angles of the triangle.
We know that exponential function \[{e^{i\theta }}\] can written as
\[{e^{i\theta }} = \cos \theta + i\sin \theta \].
We can write the following function with respect to the general form,
\[{e^{iA}} = \cos A + i\sin A\] , since \[\theta = A\] ----------(1)
\[{e^{iB}} = \cos B + i\sin B\], since \[\theta = B\] ----------(2)
\[{e^{iC}} = \cos C + i\sin C\], since \[\theta = C\] ----------(3)
Since, \[{e^{iA}}\] , \[{e^{iB}}\] , \[{e^{iC}}\] are Arithmetic Progression.
Hence, we know that the difference of two consecutive terms are equal to the A.P.
So, the difference of two consecutive terms are :
\[{e^{iB}} - {e^{iA}} = {e^{iC}} - {e^{iB}}\].
Expanding the function \[{e^{iB}}\] from RHS to LHS, we get
\[{e^{iB}} + {e^{iB}} = {e^{iA}} + {e^{iC}}\]
On simplifying we have ,
\[2{e^{iB}} = {e^{iA}} + {e^{iC}}\] ----------(4)
By substituting the equation (1), (2) and (3) in equation (4), it can be written as
\[2(\cos B + i\sin B) = (\cos A + \cos C) + i(\sin A + \sin C)\]
\[2\cos B + i(2\sin B) = (\cos A + \cos C) + i(\sin A + \sin C)\]
Comparing the reals and imaginary parts, we have
Real part: \[2\cos B = \cos A + \cos C\] ----------(5)
Imaginary part: \[2\sin B = \sin A + \sin C\] ----------(6)
On comparing this trigonometric identity with above equation,
We know that the formulas are, \[\sin A + \sin B = 2\sin \dfrac{{(A + B)}}{2}\cos \dfrac{{(A - B)}}{2}\] and \[\cos A + \cos B = 2\cos \dfrac{{(A + B)}}{2}\cos \dfrac{{(A - B)}}{2}\].
From this formula, we can write the equation (5) and (6) as follows:
Real part: \[2\cos B = \cos A + \cos C = 2\cos \dfrac{{(A + C)}}{2}\cos \dfrac{{(A - C)}}{2}\] ---------(7)
Imaginary part: \[2\sin B = \sin A + \sin C = 2\sin \dfrac{{(A + C)}}{2}\cos \dfrac{{(A - C)}}{2}\] --------(8)
On dividing the equation (8) by equation (7), then
 \[\tan B = \dfrac{{2\sin B}}{{2\cos B}} = \dfrac{{2\sin \dfrac{{(A + C)}}{2}\cos \dfrac{{(A - C)}}{2}}}{{2\cos \dfrac{{(A + C)}}{2}\cos \dfrac{{(A - C)}}{2}}}\]
On further simplifying, we can get
\[\tan B = \dfrac{{\sin \dfrac{{(A + C)}}{2}}}{{\cos \dfrac{{(A + C)}}{2}}}\]
Now, on comparing the trigonometric identity, $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ with above equation, then
\[\tan B = \tan \dfrac{{(A + C)}}{2}\]
Here, On comparing the angels we have
\[B = \dfrac{{(A + C)}}{2}\] ----------(9)
\[2B = A + C\]
Adding \[B\] on both sides, we have
\[2B + B = A + B + C\]
We know that the sum of angle of a triangle is \[180^\circ \].
\[3B = {180^ \circ }\], since \[A + B + C = {180^ \circ }\]
\[B = \dfrac{{{{180}^ \circ }}}{3}\]
Hence, \[B = 60^\circ \].
From the trigonometric degree table, $\cos {60^ \circ } = \dfrac{1}{2}$
Substitute the value \[B = 60^\circ \] in equation (5)
 \[2\cos B = \cos A + \cos C \Rightarrow \cos A + \cos C = 2\cos {60^ \circ }\]
 \[ \Rightarrow \cos A + \cos C = 2 \times \dfrac{1}{2}\] . Since, \[B = 60^\circ \]
\[ \Rightarrow \cos A + \cos C = 1\]
We use this formula, \[\cos A + \cos C = 2\cos \dfrac{{(A + C)}}{2}\cos \dfrac{{(A - C)}}{2}\].
\[ \Rightarrow 2\cos \dfrac{{(A + C)}}{2}\cos \dfrac{{(A - C)}}{2} = 1\]
\[ \Rightarrow 2\cos B\cos \dfrac{{(A - C)}}{2} = 1\].
Since, \[B = 60^\circ \Rightarrow \cos {60^ \circ } = \dfrac{1}{2}\]
\[ \Rightarrow 2 \times \dfrac{1}{2} \times \cos \dfrac{{(A - C)}}{2} = 1\]
\[ \Rightarrow \cos \dfrac{{(A - C)}}{2} = 1\]
By expanding the cosine of LHS to RHS, so it will be written as cosine inverse, then
\[ \Rightarrow \dfrac{{(A - C)}}{2} = {\cos ^{ - 1}}(1)\]
We know that ${\cos ^{ - 1}}(1) = {0^ \circ }$
\[ \Rightarrow \dfrac{{(A - C)}}{2} = 0^\circ \]
Now, we get
$A - C = 0$
Therefore, \[A = C\].
We have \[B = 60^\circ \] and \[A = C\],
Since, the sum of the three sides are equal to ${180^ \circ }$
So, $A + B + C = {180^ \circ } \Rightarrow {60^ \circ } + {60^ \circ } + {60^ \circ } = {180^ \circ }$
We can show a diagrammatic representation of the triangle as follows.
seo images

Therefore, \[A = 60^\circ \], \[B = 60^\circ \] and \[C = 60^\circ \]. So, it is an equilateral triangle.
Hence, Given that \[{e^{iA}}\] , \[{e^{iB}}\] , \[{e^{iC}}\] are in A.P. where \[A\] , \[B\], \[C\] are the angles of the triangle , then the triangle is equilateral triangle.
As a result, The option B is the correct one.
So, the correct answer is “Option B”.

Note: The terms of A.P. differ by common difference (d) and Sum of angles of the triangle is \[180^\circ \]. This is the problem of arithmetic progression.
Arithmetic progression is the branch of mathematics which deals with sequence of numbers (i.e. Terms ) whose each term differ by a common difference (denoted by d ).
For example: \[a\] , \[a + d\], \[a + 2d\], \[a + 3d\] and so on . Here each term differs by common difference.