Question

In a right-angled triangle ABC, ∠B = 90°. If BA and BC produced to the points P and Q respectively, find the value of ∠PAC + ∠QCA.

Answer 1

In ΔABC, ∠B = 90°
And, ∠ABC + ∠BAC + ∠ACB = 180°
⇒ ∠BAC + ∠ACB = 180° - ∠ABC
⇒ ∠BAC + ∠ACB = 180° - 90°
⇒ ∠BAC + ∠ACB = 90°    ....(i)

By exterior angle property,
∠PAC = ∠ABC + ∠ACB    ....(ii)
∠QCA = ∠ABC + ∠BAC  ....(iii)

Adding (ii) and (iii), we get
∠PAC + ∠QCA = ∠ABC + ∠ACB + ∠ABC + ∠BAC
⇒ ∠PAC + ∠QCA = (∠ABC + ∠BAC) + 2∠ABC
⇒ ∠PAC + ∠QCA = 90° + 2 x 90°     ....[From (i)]
⇒ ∠PAC + ∠QCA = 90° + 180°
⇒ ∠PAC + ∠QCA = 270°.