Question

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.

Answer 1

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.

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.