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In a circle, chord AB and CD meet a point P outside the circle and chord AC and BD meet a point Q outside the circle. If ∠ACD = 70° and ∠CAB = 40°, then, what is the value of (∠APC - ∠BQA)?

In a circle, chord AB and CD meet a point P outside the circle and chord AC and BD meet a point Q outside the circle. If ∠ACD = 70° and ∠CAB = 40°, then, what is the value of (∠APC - ∠BQA)? Correct Answer 40° 

Given:

∠ACD = 70° and ∠CAB = 40°

Calculation:

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We know that, ∠CAB + ∠CDB = 180°

⇒ 40° + ∠CDB = 180°

⇒ ∠CDB = 140°

Thus, ∠PDB = 180° - 140° = 40°

Similarly, ∠PBD = 70°

In ΔBPD –

∠PBD + ∠PDB + ∠BPD = 180°

⇒ 70° + 40° + ∠BPD = 180°

⇒ ∠BPD = 180° - 110°            

⇒ ∠BPD  = ∠APC = 70°       ---- (1)

In ΔCDQ –

∠CDQ + ∠DCQ + ∠DQC = 180°

⇒ 40° + (180° - 70°) + ∠DQC = 180°

⇒ ∠DQC = ∠BQA = 30°      ---- (2)

According to question -

∴ ∠APC - ∠BQA = 70° - 30° = 40°

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