Question

Please answer this question.....guys... please correct answer....

Vry important..

Don't spam!!​

Answer 1

Answer:

x = 110°

Step-by-step explanation:

∠KAY + ∠LAY = 180° (Linear Pair)

135° + ∠LAY = 180°

∠LAY = 180 - 135

∠LAY = 45°

∠LAY + ∠AYL + ∠ALY = 180° (ASP of Triangle)

45° + 35° + ∠ALY = 180°

80° + ∠ALY = 180°

∠ALY = 180 - 80

∠ALY = 100°

∠MLY + ∠ALY = 180° (Linear Pair)

∠MLY + 100° = 180°

∠MLY = 180 - 100

∠MLY = 80°

∠AYL + ∠BYL = 180° (Linear pair)

35° + ∠BYL = 180°

∠BYL = 180 - 35

∠BYL = 145°

∠BML + ∠MLY + ∠BYL + ∠MBY = 360° (ASP of Quadrilateral)

30° + 80° + 145° + ∠MBY = 360°

255° + ∠MBY = 360°

∠MBY = 360 - 255

∠MBY = 105°

∠BYD = ∠AYL = 35° (Vertically Opposite Angle)

∠MBY + ∠DBY = 180° (Linear Pair)

105° + ∠DBY = 180°

∠DBY = 180 - 105

∠DBY = 75°

∠BYD + ∠DBY + ∠BDY = 180° (ASP of Triangle)

35° + 75° + ∠BDY = 180°

110° + ∠BDY = 180°

∠BDY = 180 - 110

∠BDY = 70°

∠BDY + x = 180° (Linear Pair)

70° + x = 180°

x = 180 - 70

x = 110°

Alternate method after solving ∠DBY

x is exterior angle of ▲BDY, so

x = ∠BYD + ∠DBY (Exterior Angle Property)

x = 35° + 75°

x = 110°

Hope it helps...

Plz mark me as brainliest.