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In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75o, ∠ABD = 58o and ∠ADC = 77o , AC and BD intersect at P. the measure of ∠DPC is



  • a)
    92o

  • b)
    105o

  • c)
    90o

  • d)
    94o

Correct answer is option 'A'. Can you explain this answer?
Verified Answer
In the given figure, ABCD is a cyclic quadrilateral in which∠BAD =...
∠DBA = ∠DCA = 58° …(1)
[Angles in same segment]  ABCD is a cyclic quadrilateral : 
Sum of opposite angles = 180 degrees 
∠A +∠C = 180° 
75° + ∠C = 180° 
∠C = 105° 
Again, 
∠ACB + ∠ACD = 105° 
∠ACB + 58° = 105° 
or ∠ACB = 47° …(2) 
Now,  ∠ACB = ∠ADB = 47° [Angles in same segment] 
Also, ∠D = 77° (Given) 
Again From figure, ∠BDC + ∠ADB = 77° 
∠BDC + 47° = 77° 
∠BDC = 30° 
In triangle DPC 
∠PDC + ∠DCP + ∠DPC = 180° 
30° + 58° + ∠DPC = 180° 
or ∠DPC = 92° 
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Most Upvoted Answer
In the given figure, ABCD is a cyclic quadrilateral in which∠BAD =...
ANS:- let ∠PBC = x

∠ADC + ∠ABC = 180... ( opposite angles of a cyclic quadrilateral are supplementary)

77+ ∠ABD + x =180.. (∠ABC =∠ABD + x)

77+58+x =180.
135 + x =180
x = 180-135
x =45
∠PBC=45..(1)

Now , in ∆ADB,

∠ABD+∠BAD+ ∠ADB=180

58+75+ ∠ADB=180.

133+∠ADB=180.

∠ADB=180-133

∠ADB=47.

Here,. ∠ACB = ∠ADB... (angles in the same segment are equal)

Therefore,. ∠ACB=47....(2)

Now in ∆DPC,
∠DPC = ∠ACB + ∠PBC ..(exterior angle theoram.)

∠DPC= 47+45.... [from(1) and (2)]

Therefore,. ∠DPC=92.
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Community Answer
In the given figure, ABCD is a cyclic quadrilateral in which∠BAD =...
A is correct
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