In a ΔABC, points P, Q and R are taken on AB, BC and CA, respectively, such that BQ = PQ and QC = QR. If ∠BAC = 75º, what is the measure of ∠PQR (in degrees)?

Shortcut Trick

∠BAC = 75º

∠ABC + ∠ACB + ∠BAC = 180°

∠ABC + ∠ACB + 75° = 180°

∠ABC + ∠ACB = 180° - 75° = 105°

Let, ∠ABC = ∠PBQ = 70° and ∠ACB = ∠RCQ = 35°

So, ∠PQR = 180° - (∠PQB + ∠RQC)

= 180° - [(180° - 2∠PBQ) + (180° - 2∠RCQ)  [∵ BQ = PQ; QC = QR]

= 180° - [(180° - 2 × 70°) + (180° - 2 × 35°)]

= 180° - (40° + 110°)

= 180° - 150°

= 30°

Alternate Method

Given:

In a ΔABC, ∠BAC = 75º

BQ = PQ and QC = QR

Concept used:

The sum of all three angles of a triangle = 180°

The sum of all angles on a straight line = 180°

Calculation:

Let, ∠ABC = x and ∠ACB = y

So, ∠ABC = ∠PBQ = ∠QPB = x  [∵ BQ = PQ]

∠ACB = ∠RCQ = ∠QRC = y  [QC = QR]

In ΔABC, ∠ABC + ∠ACB + ∠BAC = 180°

⇒ x + y + 75° = 180°

⇒ x + y = 180° - 75° = 105°   .....(1)

For ΔBPQ and ΔCRQ,

(∠PBQ + ∠QPB + ∠PQB) + (∠RCQ + ∠QRC + ∠RQC) = 180° + 180° = 360°

⇒ (x + x + ∠PQB) + (y + y + ∠RQC) = 360°

⇒ 2x + 2y + ∠PQB + ∠RQC = 360°

⇒ 2 (x + y) + ∠PQB + ∠RQC = 360°

⇒ (2 × 105°) + ∠PQB + ∠RQC = 360°  [∵ x + y = 105°]

⇒ ∠PQB + ∠RQC = 360° - 210° = 150°   .....(2)

Also, ∠PQB + ∠RQC + ∠PQR = 180°

⇒ 150° + ∠PQR = 180°  [∵ ∠PQB + ∠RQC = 150°]

⇒ ∠PQR = 180° - 150° = 30°

∴ The measure of ∠PQR (in degrees) is 30°