Answer:
Explanation:
Given :
In triangle ABC, ∠A = 32°,
∠B = 49°
In triangle A'B'C',
∠B' = 49°
∠C' = 99°
To prove : ΔABC and ΔA'B'C' are similar.
Proof :
In triangle ABC,
∠A = 32° and ∠B = 49°
Then ∠C = 180° - (∠A + ∠B)
= 180° - (32° + 49°)
= 180° - 81°
= 99°
Similarly, in ΔA'B'C'
∠B' = 49° and ∠C' = 99°
Then ∠A' = 180° - (99° + 49°)
= 180° - 148°
= 32°
Now we find ∠A ≅ ∠A' ≅ 32°
∠B ≅ ∠B' ≅ 49°
and ∠C ≅ ∠C' ≅ 99°
Therefor, by both the triangles ABC and A'B'C' will be similar.