Question

Abcd is a cyclic quadrilateral and pq is a tangent to the circle at c. If bd is a diameter dcq is 40 and abd is 60 then find dbc bcp bdc adb.

Answer 1

DAB = 90° (Angle in a semi-circle)

In ΔABD,

∠ABD + ∠ADB + ∠DAB = 180°

⇒ ∠ADB = 180° – (60° + 90°)

⇒∠ADB = 180° – 150°

⇒∠ADB = 30°

∠DCQ = ∠CBD = 40°

(Alternate segment theorem)

∠BCD = 90°

In ΔBCD,

∠BCD + ∠BDC + ∠DBC = 180°

⇒ ∠BDC = 180° – (90° + 40°) = 50°

∠BDC = ∠BCP = 50° (Alternate segment theorem)

∴ (i) ∠ADB = 30°

(ii) ∠BCP = 50°

Answer 2

Answer:

We are given that ABCD is a cyclic quadrilateral which means sum of opposite angles is 90°

Angle(DAB+BCD) =180°

Angle(CDA+ABC)=180°

Angle(QCD+DCB+BCP)=180° - linear pair

Angle(40°+90°+BCP)=180°

Angle BCP=50°[180-130]

Angle CDB=PCB=50°(alternate angle Theorem)

Angle dab + abd + bda = 180 (angle sum prop. Of triangle)

90+60+bda=180

Bda =30