Concept:We use the trigonometric identity that relates secant and cosecant: sec θ = cosec (90° – θ). This allows us to convert sec 6A into a cosecant expression and then equate the angles.
Explanation:Given: sec 6A = cosec (A – 29°) and 2A is an acute angle.
Rewrite sec 6A using the identity: sec 6A = cosec (90° – 6A).
Now the equation becomes: cosec (90° – 6A) = cosec (A – 29°).
Since cosecant is positive for acute angles and the arguments are likely acute, we equate the angles: 90° – 6A = A – 29°.
Solve step by step: 90° – 6A = A – 29° → bring terms together: 90° + 29° = A + 6A → 119° = 7A.
Divide both sides by 7: A = 119° / 7 = 17°.
Check: If A = 17°, then 2A = 34° (acute) and A – 29° = –12°? Wait, cosec of negative angle? Note: cosec(–12°) = –cosec(12°) but sec(6×17°)=sec(102°) which is negative? Actually 6A=102°, sec 102° is negative, cosec(–12°) is also negative. However the condition "2A is acute" ensures the identity holds for conversion? The original solution ignores signs because the identity sec θ = cosec(90°–θ) holds for all θ where defined. So A=17° is correct.
Answer:∠A = 17º, which corresponds to option C.