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please help me with these questions
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Answer:
1. y = 80
2. x = 110
Step-by-step explanation:
1. ∠PRS and 135° are supplementary
∠PRS = 180-135=45
∠QRS = 55° + ∠PRS = (55 + 45)° = 100°
∠P + ∠Q + ∠QRS + ∠S = 360° (Using angle sum property)
⇒ (90 + 90 + 100 + ∠Y)° = 360° (Given)
⇒ ∠Y + 280 = 360
⇒ ∠Y = 360 - 280 = 80
∴ ∠Y = 80°
2. ∠ABC = 105° (Using angle sum property)
∠CBD = 75° (Supplementary angles)
∠X = 35 + 75 (External angle property)
= 110°
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Verified answer
Answer:
1) Y= 80 °, 2) 110°
Step-by-step explanation:
ANS 1.) In ∆ RPQ
angle RQP + PRQ + RPQ = 180°
90° + 55° + angle RPQ = 180°
angle RPQ = 180° - ( 90° + 55° )
angle RPQ = 45°
IN ∆ SPR
angle SPR + RPQ = 90° ( GIVEN )
angle SPR = 90° - RPQ = 45°
angle SPR + Y = 135° ( EXTERIOR ANGLE SUM PROPERTY)
angle Y = 135° - 45° = 80°
ANS 2.) IN ∆ ABC
angle ACB + angle BAC + angle ABC = 180° ( SUM OF INTERIOR ANGLES OF ∆ )
25° + 50° + angle ABC = 180°
angle ABC = 180° - 75° = 105°
Now in ∆BDE
angle EBD + ABC = 180° ( linear pair )
angle EBD = 75°
angle x = angle EDB + EBD
x = 75° + 35° = 110°
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