Answer:
A = 40°
Step-by-step explanation:
[tex] \sin 3A = \cos (A – 70ᵒ) \\ \\ \cos(90 \degree - 3A) = \cos (A – 70ᵒ) \\ \{\because \: \cos(90 \degree - \theta) = \sin \theta \} \\ \\ (90 \degree - 3A) = (A – 70ᵒ) \\ \\ 90 \degree + 70ᵒ = 3A + A \\ \\ 160 \degree = 4A \\ \\ A = \frac{160 \degree}{4} \\ \\ \huge \red{ \boxed{A = 40 \degree}}[/tex]
