In Fig.6.3, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠ APB = 50° and ∠ CDP = 30°. Then, ∠ PBA is equal to

(A) 50°  (B) 30°   (C) 60°   (D) 100°

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  1. Chapter 6 Class 10 Triangles (Term 1)
  2. Serial order wise

Transcript

Question 7 In Fig.6.3, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠ APB = 50° and ∠ CDP = 30°. Then, ∠ PBA is equal to (A) 50° (B) 30° (C) 60° (D) 100° In Δ APB and Δ DPC 𝐴𝑃/𝐵𝑃=𝐷𝑃/𝐶𝑃 ∠ APB = ∠ DPC ∴ Δ APB ~ Δ DPC Since both triangles are similar Their angles will be equal ∴ ∠ PAB = ∠ PDC = 30° Now, In Δ APB, ∠ ABP + ∠ PAB + ∠ APB = 180° ∠ ABP + 30° + 50° = 180° ∠ ABP + 80° = 180° ∠ ABP = 180° − 80° ∠ ABP = 100° So, the correct answer is (D)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.