$\text{Ta có: ∠A + ∠B + ∠C = $180^{o}$ ( tổng 3 góc trong Δ )}$
$\text{ mà : ∠C = $14^{o}$ }$
$\text{⇒ ∠A + ∠B = $166^{o}$}$
$\text{mà ∠A = 3.∠B}$
$\text{}$
$\text{⇒ 3.∠B + ∠B = $166^{o}$ }$
$\text{⇒ 4∠B = $166^{o}$}$
$\text{⇒ ∠B = $41,5^{o}$}$
$\text{Ta có : ∠A = 3.∠B}$
$\text{mà ∠B = $41,5^{o}$}$
$\text{⇒ ∠A = $124,5^{o}$}$