Question

10: If sin 3A= cos (A– 26°), where 3A is an acute angle, find the value of​

Answer 1

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$sin \: 3a = cos(a \: - \: 26) \\ \\ sin \: a = cos(90 - a) \\ \\ hence \\ \\ cos(90 - 3a) = cos(a - 26) \\ \\ 90 - 3a = a - 26 \\ \\ 4a = 116 \\ \\ a = \frac{116}{4} \\ \\ a = 29$

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Answer 2

$\huge {\purple{\boxed{ \underline{ \underline{ \bigstar{\red{answer}}}}}}}$

$\huge {\green {\underline {\underline{given}}}} = >$

$\sin3 \ A = \cos(A - {26}^{0} )$

$\huge {\green {\underline {\underline{solution}}}} = >$

${ \boxed{ \sin \: A = \cos(90 -A )}}$

so,

$\ \cos (90 - 3A) = \cos( A - 26)$

$90 - 3A = A - 26 \\ \\ { \huge{90 + 26 = A + 3A}} \\ \\ 116 = 4A \\ \\ {\huge{A = \frac{116}{4} }} \\ \\ { \huge{A = 29}}$

$\huge { \underline{ \boxed{ \pink{answer = >A = 29 }}}}$

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