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amit

· started a discussion

· 1 Months ago

wrong answer ..ans is100

Question:
\(\triangle\)BCD is a cycle triangle. AD is tangent to the circle from a point A outside the circle. Side BC is subtended upto point P such that point A lies between point B and P and points C, B, A and P are in straight line. Given \(\angle \)PAD = 140°, AB = BD and O is the centre of circle. Find the value of \(\angle \)DBC. 
Options:
A) 60°
B) 120°
C) 80°
D) 100°
Solution:
Ans: (c) 


Shubham kumar

· commented

· 1 Months ago

Yet to be approved!

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