Question

In fig. QT perpendicular PR & Qs =ps. If
measure of angle TQR = 40° & RPS = 20, then
the value of x is...
​

Answer 1

Answer:

Step-by-step explanation:

n Δ QTR  

∠ TQR + ∠ QRT + ∠ QTR = 180°

⇒ 40° + y + 90° = 180°

⇒ y = 180° - 130°

⇒ y = 50°

∠ QSP = ∠ SPR + ∠ SRP

Reason : Exterior angle = sum of interior angles

⇒ x = 30° + y

⇒ x = 30° + 50°

⇒ x = 80°

Answer 2

Answer:

X= 15°

Step-by-step explanation:

Given PS = QS

=> /_SPQ = x+40 .....(1) (Angles opposite to equal sides are equal)

Now QT is perpendicular to PR

This means Angle QTR=QTP = 90°

Therefore in Triangle QTR,

40°+90°+Angle TRQ = 180° (Angle sum property)

or Angle TRQ = 50°.....(2)

Therefore, by angle sum property in Triangle PSR, Angle PSR =110°

This implies Angle PSQ = 70° (linear pair)

Now, let's assume the point where QT and PS intersect be O.

Therefore, Angle POQ = 180°- (180°-70°-40°) (Linear Pair)

This implies Angle POQ = 110°.....(3)

Now, finally in Triangle PSQ

Angle SPQ = x+40 (from 1)

This implies x+40 = 180°-110°-x (Angle sum property in Triangle POQ)

or 2x = 30°

or x = 15°.