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vishwanath tripathi

· started a discussion

· 1 Months ago

counting angle (Question: 113112.0 )

please make sure all angle measure are correct

Question:
The sides AB and CD of a cyclic quadrilateral ABCD are produced to meet at P, the sides AD and BC are produced to meet at Q. If \(\angle \)ADC = 85°, \(\angle \)BPC = 40° and \(\angle \)CBP = 85°, then \(\angle \)CQD equals :
Options:
A) \(30\ ^o\)
B) \(45\ ^o\)
C) \(60\ ^o\)
D) \(75\ ^o\)
Solution:
Ans: (a) \(\angle \)ABC + \(\angle \)ADC = 180° 


95° + \(\angle \)ADC = 180°

\(\angle \)ADC = 85°............... (i)

\(\angle \)ABC + CBP = 180°

95 + \(\angle \)CBP = 180°

\(\angle \)CBP = 85°................ (ii)

In \(\triangle\)BCP

\(\angle \)CBP + \(\angle \)BPC + \(\angle \)PCB = 180°

85° + 40° + \(\angle \)PCB = 180°

PCB = 55° = \(\angle \)QCD

\(\angle \)ADC + \(\angle \)QDC = 180°

85° + \(\angle \)QDC = 180°

\(\angle \)QDC = 180° - 85° = 95°

In \(\triangle\)QDC

\(\angle \)QDC + \(\angle \)DQC + \(\angle \)QCD = 180°

95° + \(\angle \)DQC + 55° = 180°

\(\angle \)DQC = 30°

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