ABCD is a cyclic quadrilateral, AB = AD.
PAQ is a tangent to the circle.
BA is produced to a point T.
ÐDAT = 112°, Find
(i) ÐPAD (ii) ÐAOB
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Answer:
Join OC,OD and AC
(i)
∠BCG+∠BCD=180
o
[Linear pair]
⇒108
o
+∠BCD=180
o
⇒∠BCD=180
o
−108
o
=72
o
BC=CD
∴∠DCP=∠BCT
But,∠BCT+∠BCD+∠DCP=180
o
∴∠BCT+∠BCT+72
o
=180
o
2∠BCT=180
o
−72
o
∠BCT=54
o
(ii)
PCT is tangent and CA is a chord
∠CAD=∠BCT=54
o
But arc DC subtends ∠DOC at the centre and ∠CAD at the remaining part of the circle.
∴∠DOC=2∠CAD=2×54
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