Question

If A,B,C are angles of triangle such that x = cisA, y = cisB, z = cisC then find the value of x.y.z

Answer 1

$\mathrm\red{Given : }$

$\mathrm{A, \: B, \: C \: are \: the \: angles \: of \: triangle}$

$\mathrm{A + B + C = 180°}$

$\mathrm\red{Also \: given : \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \mathsf{x = cis \: A, \: y = cis \: B, \: z = cis \: C } \\ \\ \mathrm{xyz \: = cis \: A.cis \: B. cis \: C \: \: \: \: \: \: \: \: \: } \\ \\ \mathrm{ = \: cis(A + B + C)}$

$\mathrm{ = \cos(A+B+C) \: + \: i \sin(A+B+C)} \\ \\ \mathrm{ = \: \cos \: 180 \: + \: i \sin 180} \\ \\ \mathrm{ = \: - 1 + 0} \\ \\ \mathrm{ = - 1}$

$\boxed{ \mathrm \red{ \therefore \: \: xyz \: = \: - 1}}$

Answer 2

Step-by-step explanation:

$\begin{gathered} \mathrm\red{Given : } \\ \end{gathered}$

$\begin{gathered} \mathrm{A, \: B, \: C \: are \: the \: angles \: of \: triangle} \\ \end{gathered}$

$\begin{gathered} \mathrm{A + B + C = 180°} \\ \end{gathered}$

$\begin{gathered} \mathrm\red{Also \: given : \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \\ \mathsf{x = cis \: A, \: y = cis \: B, \: z = cis \: C } \\ \\ \mathrm{xyz \: = cis \: A.cis \: B. cis \: C \: \: \: \: \: \: \: \: \: } \\ \\ \mathrm{ = \: cis(A + B + C)} \\ \end{gathered}$

$\begin{gathered} \mathrm{ = \cos(A+B+C) \: + \: i \sin(A+B+C)} \\ \\ \mathrm{ = \: \cos \: 180 \: + \: i \sin 180} \\ \\ \mathrm{ = \: - 1 + 0} \\ \\ \mathrm{ = - 1} \\ \\ \end{gathered}$

$\boxed{ \mathrm \red{ \therefore \: \: xyz \: = \: - 1}}$