If cos(40+x)° = sin 30° find the valueof x

1.	If cos(40+x)° = sin 30° find the valueof x
Answer» $\begin{gathered}\begin{gathered}\bf \: <klux>GIVEN</klux> - \begin{cases} &\sf{cos(40 + x)\degree \: = sin30\degree \:} \\ \end{cases}\end{gathered}\end{gathered}$ $\begin{gathered}\begin{gathered}\bf \:To\:find - \begin{cases} &\sf{the \: value \: of \: x} \\ \end{cases}\end{gathered}\end{gathered}$ $\begin{gathered}\Large{\sf{{\underline{Formula \: Used - }}}} \end{gathered}$ $\boxed{ \bf{sin(90\degree \: - x) = cosx}}$ $\large\underline{\sf{Solution-}}$ Given that, $\rm :\longmapsto\:cos(40 + x)\degree \: = sin30\degree \:$ $\rm :\longmapsto\:cos(40 + x)\degree \: = sin(90\degree \: - 60\degree)$ $\rm :\longmapsto\:cos(40 + x)\degree \: = cos60\degree \:$ So, on COMPARING, we GET $\rm :\longmapsto\:40 + x = 60$ $\bf\implies \:x \: = \: 20$ ADDITIONAL INFORMATION :- $1. \: \: \boxed{ \bf{sin(90\degree \: - x) = cosx}}$ $2. \: \: \boxed{ \bf{cos(90\degree \: - x) = sinx}}$ $3. \: \: \boxed{ \bf{tan(90\degree \: - x) = cotx}}$ $4. \: \: \boxed{ \bf{cot(90\degree \: - x) = tanx}}$ $5. \: \: \boxed{ \bf{sec(90\degree \: - x) = cosecx}}$ $6. \: \: \boxed{ \bf{cosec(90\degree \: - x) = secx}}$

Answer»

$\begin{gathered}\begin{gathered}\bf \: <klux>GIVEN</klux> - \begin{cases} &\sf{cos(40 + x)\degree \: = sin30\degree \:} \\ \end{cases}\end{gathered}\end{gathered}$