Question

If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.

Answer 1

$$\begin{gathered} \begin{array}{l}\text{sin3A}=\cos (A-{26}^0)\\ \end{array} \\ \begin{array}{l}\text{⇒cos}({90}^0-3A)=\cos (A-{26}^0)[\because \sin \theta =\cos ({90}^0-\theta )]\\ \end{array} \\ \begin{array}{l}\Rightarrow {\text{90}}^0-3A=A-{26}^0\\ \end{array} \\ \begin{array}{l}\Rightarrow {\text{116}}^0=4A\\ \end{array} \\ \Rightarrow A=\frac{{116}^0}{4}={29}^0 \end{gathered}$$