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प्रश्न
If sin 3 A = cos (A − 26°), where 3 A is an acute angle, find the value of A.
उत्तर
\[\begin{array}{l}sin3A = \cos(A =- {26}^\circ ) \\ \end{array}\]
\[\begin{array}{l}\Rightarrow cos( {90}^\circ - 3A) = \cos(A-{26}^\circ )[ \because \sin\theta = \cos( {90}^\circ - \theta)] \\ \end{array}\]
\[\begin{array}{l}\Rightarrow {90}^\circ - 3A = A - {26}^\circ \\ \end{array}\]
\[\begin{array}{l}\Rightarrow {116}^\circ = 4A \\ \end{array}\]
\[ \Rightarrow A = \frac{{116}^\circ}{4} = {29}^\circ \]
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