We know that, in a triangle , the sum of three angles is 180°
∴ ∠A + ∠B + ∠C = 180° …(a)
According to the question,
On taking II and III, we get
⇒ 3∠B = 2 (∠A + ∠B)
⇒ 3∠B = 2 ∠A + 2 ∠B
⇒ ∠B = 2 ∠A …(i)
Now, on taking I and II, we get
∠C = 3 ∠B
⇒ ∠C = 3(2 ∠A) (from eq. (i))
⇒ ∠C = 6 ∠A …(ii)
On substituting the value of ∠B and ∠C in Eq. (a), we get
∠A + 2∠A + 6∠A = 180°
⇒ 9∠A = 180°
⇒ ∠A = 20°
On putting the value of ∠A = 20° in Eq. (i) and (ii), we get
∠B = 2 ∠A = 2(20) = 40°
∠C = 6 ∠A = 6(20) = 120°
Hence, the angles are ∠A = 20°, ∠B = 40°, ∠C = 120°
