Question

In a ΔABC, AB = AC. P and Q are points on AC and AB respectively such that CB = BP = PQ = QA. Then ∠AQP =

(a) \(\frac{2π}{7}\)

(b) 3π 

(c) \(\frac{5π}{7}\)

(d) \(\frac{4π}{7}\)

Answer 1

(c) \(\frac{5π}{7}\)

Let ∠AQP = α 

In ΔAQP, AQ = QP 

⇒ ∠QAP = ∠QPA = \(\frac{1}{2}\) (π - α)

= \(\frac{π}{2}\) - \(\frac{α}{2}\)

∠PQB = π - α       (AQB being a straight line)

In ΔPQB, 

PQ = PB ⇒ ∠PBQ = ∠PQB = π - α 

∴ ∠QPB = π - 2 (π - α) = 2α – π

In ΔABC, 

∠BAC + ∠ABC + ∠ACB = π 

⇒ ∠QPA + ∠ACB + ∠ACB = π 

(∵ ∠QAP = ∠BAC, AB = AC ⇒ ∠ACB = ∠ABC)

⇒ \(\frac{π}{2}\) - \(\frac{α}{2}\) + 2∠ACB = π

⇒ 2∠ACB = π - \(\frac{π}{2}\) + \(\frac{α}{2}\) = \(\frac{π}{2}\) + \(\frac{α}{2}\) ⇒ ∠ACB = \(\frac{π}{4}\) + \(\frac{α}{4}\)

 ∴ In ΔBPC, BP = BC 

⇒ ∠BPC = ∠BCP = ∠ACB = \(\frac{π}{4}\) + \(\frac{α}{4}\)

Now APC being a straight line, 

∠APQ + ∠BPQ + ∠BPC = π

⇒ \(\frac{π}{2}\) - \(\frac{α}{2}\) + 2α - π + \(\frac{π}{4}\) + \(\frac{α}{4}\)= π

⇒ \(\frac{7α}{4}\) = \(\frac{5π}{4}\) ⇒ α = \(\frac{5π}{7}\).