Assertion (A): In ΔABC and ΔPQR, if ∠BAC = ∠QPR and ∠ABC = ∠PQR, then ΔABC ~ ΔPQR

Statement 2: ΔABC ~ ΔPQR by SSS axiom

1. (A) is true, (R) is false.

2. (A) is false, (R) is true.

3. Both (A) and (R) are true, and (R) is the correct reason for (A).

4. Both (A) and (R) are true, and (R) is the incorrect reason for (A).

In ΔABC and ΔPQR,

⇒ ∠BAC = ∠QPR [Given]

⇒ ∠ABC = ∠PQR [Given]

∴ ΔABC ~ ΔPQR (By A.A. axiom)

So, assertion (A) is true and reason (R) is false.

Hence, option 1 is the correct option.