Question

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

Answer 1

SIN 3A =COS(A-26)

SIN 3A=SIN(90-(A-26))

SIN 3A=SIN(90-A+26)

Sin 3A=SIN(116-A)

SIN SIN CANCELLED

SO, 3A=116-A

3A-A=116

2A=116

A=116/2

A=58ans

I hope it is helpful ☺️

Answer 2

Answer:

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