Question

KLMN is an isosceles trapezium whose diagonals cut at X and KL is parallel to NM. If ∠KNL = 25°, ∠KMN = 30°, find ∠KXN and ∠MLN
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Answer 1

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KN = LM (Isos.trap.)

∠NKL = ∠KLM ( Isosceles trapezium property)

KL = KL (common)

∴ ΔKLN ≅ Δ KLM (S.A.S. axiom)

⇒ ∠KNL = ∠KML = 25°

∴ ∠LMN = 25° + 30°

= 55°

Now,

∠KNM = ∠LMN

⇒ ∠KNL + ∠LNM = 25°

⇒ 25° + ∠LMN = 55°

⇒ ∠LNM = 55° -25°

⇒ ∠LNM = 30°

In ΔNXM,

∠NXM = 180°-(∠XNM + ∠XMN)

= 180° - (30°+30°)

= 180° - 60°

= 120°

∴ ∠KXN = 180° - ∠NXM (Linear pair)

= 180° - 120°

= 60°

∴ ∠KXN measures 60°

∠LXM = ∠KXN = 60° (vertically opposite angles)

Now,

In ΔMLX,

∠MLX = 180° - (60°+25°)

= 180° - 85°

= 95°

∴ ∠MLX measures 95°

Answer 2

Answer:In Δs KLN and KLM,

KN = LM (Isos.trap.)

∠NKL = ∠KLM ( Isosceles trapezium property)

KL = KL (common)

∴ ΔKLN ≅ Δ KLM (S.A.S. axiom)

⇒ ∠KNL = ∠KML = 25°

∴ ∠LMN = 25° + 30°

             = 55°

Now,

∠KNM = ∠LMN

⇒ ∠KNL + ∠LNM = 25°

⇒ 25° + ∠LMN = 55°

⇒ ∠LNM =  55° -25°

⇒ ∠LNM = 30°

In ΔNXM,

∠NXM = 180°-(∠XNM + ∠XMN)

           = 180° - (30°+30°)

           = 180° - 60°

           = 120°

∴ ∠KXN = 180° - ∠NXM (Linear pair)

               = 180° - 120°

              = 60°

∴ ∠KXN measures 60°

∠LXM = ∠KXN = 60° (vertically opposite angles)

Now,

In ΔMLX,

∠MLX = 180° - (60°+25°)

          = 180° - 85°

          = 95°

∴ ∠MLX measures 95°

Step-by-step explanation: