Question

In the given figure, PQR is a triangle, PS ⊥ QR and PT is bisector of ∠QPR that meets QR at T. PSV and PTW are rays such that U is the intersection point of the angle bisectors of the exterior angles TSV and STW. If ∠Q = 80° and ∠R = 40°, then the difference between ∠SPT and ∠SUT is

• A. 40°
• B. 60°
• C. 80°
• D. 90°

Answer 1

The correct option is B 60°

Given: ∠Q = 80°, ∠R = 40°



In ΔPQR, by Angle sum property,
∠PQR + ∠PRQ + ∠QPR = 180°
⇒ 80° + 40° + ∠QPR = 180°
⇒ ∠QPR = 60°
⇒ ∠QPT = ∠TPR = 30° (∵ PT bisects ∠QPR) …..(i)
Now in ΔQPS, by Angle sum property,
∠QPS + ∠PQS + ∠PSQ = 180°
⇒ ∠QPS = 180° – (90° + 80°) (∵ PS ⊥ QR)
⇒ ∠QPS = 10° …..(ii)
∴ ∠SPT = ∠QPT – ∠QPS
⇒ ∠SPT = 30° – 10° [From (i) and (ii)]
⇒ ∠SPT = 20° …..(iii)
Here, ∠TSV = 90°. (∵ PS ⊥ QT)
$\therefore \angle TSU=\frac{{90}^{\circ }}{2}$ (∵ SU bisects ∠TSV)
⇒ ∠TSU = 45° …..(iv)
Now, ∠STW + ∠STP = 180° (Linear pair)
⇒ ∠STW = 180° – 70° = 110°
$\Rightarrow \angle STU=\frac{{110}^{\circ }}{2}$ (∵ TU bisects ∠STW)
⇒ ∠STU = 55° …..(v)
In ΔSTU, by Angle sum property,
∠STU + ∠TSU + ∠SUT = 180°
⇒ ∠SUT = 180° – (45° + 55°) = 180° – 100° [From (iv) and (v)]
⇒ ∠SUT = 80° …..(vi)
Now, ∠SUT – ∠SPT = 80° – 20° [From (iii) and (iv)]
= 60°
Hence, the correct answer is option (b).