Question

11. ABCD is a cyclic quadrilateral and PQ is a tangent at A to the circle
circumscribing the quadrilateral. If BD is a dimeter, ZABD = 30° and
ZBDC=60° then find -
(i) ZQAD
(ii) ZBAD (iii) ZPAB
(iv) ZBCD and
(v) ZCBD.​

Answer 1

Answer:

Since BD is a diameter of the circle.

∴∠BAD=90

∘

and also ∠BCD=90

∘

∠DBC=∠DCQ=40

∘

[∠s in the alternate segments]

∠BCP+∠BCD+∠DCQ=180

∘

⇒∠BCP+90

∘

+40

∘

=180

∘

∠BCP=50

∘

Similarly from ΔBAD,60

∘

+∠BAD+∠ADB=180

∘

⇒60

∘

+90

∘

+∠ADB=180

∘

∠ADB=30

∘

.