In the diagram, CDP is a straight line, ΔAQD is equilateral ∠BAR= 9

1.	In the diagram, CDP is a straight line, ΔAQD is equilateral ∠BAR= 90°, ∠QAR= 135°, ∠BCD = 106° and ∠ABC = 100°. Then, ∠PDQ equals.(a) 39° (b) 21° (c) 41° (d) 53°
Answer» (c) 41° ΔAQD is an equilateral Δ ⇒ ∠QAD = ∠QDA = ∠AQD = 60° ∠BAD = 360° – (135° + 90° + 60°) = 360° – 285° = 75° (Angles round a pt.) Also, in quad. ABCD, ∠CDA = 360°– (100° + 106° + 75°) = 360° – 281° = 79°. ∴ ∠PDQ =180° – (∠CDA+ ∠QDA) (CDP is a st. line) = 180° – (79° + 60°) = 180° – 139° = 41°.

In the diagram, CDP is a straight line, ΔAQD is equilateral ∠BAR= 90°, ∠QAR= 135°, ∠BCD = 106° and ∠ABC = 100°. Then, ∠PDQ equals.(a) 39° (b) 21° (c) 41° (d) 53°

Answer»

(c) 41°

ΔAQD is an equilateral Δ

⇒ ∠QAD = ∠QDA = ∠AQD = 60°

∠BAD = 360° – (135° + 90° + 60°)

= 360° – 285° = 75° (Angles round a pt.)

Also, in quad. ABCD,

∠CDA = 360°– (100° + 106° + 75°)

= 360° – 281° = 79°.

∴ ∠PDQ =180° – (∠CDA+ ∠QDA) (CDP is a st. line)

= 180° – (79° + 60°) = 180° – 139° = 41°.