Question

sinA=cos(A-26) where 3a is sin acute angle find the value of A​

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$Given, \: \sin(3A) = \cos(A - {26}^{o} ) \: \: \: \: \: \: ...........(i) \\ where, \: 3A \: is \: an \: acute \: angle. \\ We \: know \: that, \: \: sin ϑ  = cos( {90}^{o} -  ϑ ) \\ From \: Eq. \: (i), \: cos( {90}^{o} - 3A) = cos(A - 26) \\ Since, \: ( {90}^{o} - 3A) \: and \: (A - {26}^{o} ) \: both \: are \: acute \: angles. \\ ∴ \: {90}^{o} -3A = A - {26}^{o} \\ ➪ \: 4A = {116}^{o} \\ ➪ \: A = \frac{ {116}^{o} }{4} \\ ➪ \: A = {29}^{o}$