Question

if sinA=cos(A-26°) where 3A is acute angle find the value of A

Answer 1

Answer:

Step-by-step explanation:

Sin3A=cos(A-26) Sin3A=sin26 3A=26 A=26÷3

Solution : We are given that sin 3A = cos (A – 26°). ---------(1)

Since sin 3A = cos (90° – 3A), we can write (1) as

cos (90° – 3A) = cos (A – 26°)

Since 90° – 3A and A – 26° are both acute angles, therefore,

90° – 3A = A – 26°

which gives A = 29°

Answer 2

Step-by-step explanation:

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