Question

Find the value of A if sec 3A = cosec (A – 26°) where 3A and (A – 26°) are acute angles.

Answer 1

Given: sec 3A = cosec (A – 26°), with 3A and (A – 26°) acute angles.

Step-wise calculation:

1. sec 3A = cosec (A – 26°)

⇒ `1/(cos 3A) = 1/(sin(A - 26^circ))` 

⇒ cos 3A = sin (A – 26°)

2. Write sin in terms of cosine:

sin (A – 26°) = cos [90° – (A – 26°)]

= cos (116° – A) 

So cos 3A = cos (116° – A).

3. Equate angles for cosine:

Either (i) 3A = 116° – A + 360°k

⇒ 4A = 116° + 360°k

⇒ A = 29° + 90°k

or (ii) 3A = –(116° – A) + 360°k 

⇒ 3A = A – 116° + 360°k

⇒ 2A = –116° + 360°k 

⇒ A = –58° + 180°k, where k is any integer.

4. Apply the acute-angle constraints:

0° < 3A < 90°

⇒ 0° < A < 30° 

0° < A – 26° < 90°

⇒ 26° < A < 116° 

Intersection: 26° < A < 30°.

5. From the candidate solutions, only A = 29° from A = 29° + 90°k with k = 0 lies in (26°, 30°). The other families do not satisfy the intersection.

A = 29°.