Question

in quadrilateral ABCD is cyclic if m(arc BC) =90° and angle DBC=55° . Thaen find the measure of angle BCD​

Answer 1

Answer:

m(arc BC)=90°   ⇒(given)

∴m∠BDC= $\frac{1}{2\\}$ m(arc BC)   ⇒(Inscribed Angle Theorem)

∴m∠BDC= $\frac{1}{2\\}$ ×90°

∴m∠BDC=45°  ⇒(1)

Now, Consider ΔBDC

         m∠BDC + m∠BCD + m∠DBC = 180°  ⇒( Angle sum property of Triangle )

But, m∠BDC=45°  ⇒(from 1)

       m∠DBC = 55°  ⇒ (given)

∴ 45° +  m∠BCD + 55° = 180°

∴  m∠DCB + 100° = 180°

           ∴ m∠BCD = 180° - 100°

           ∴ m∠BCD = 80°

Step-by-step explanation:

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Answer 2

Given : quadrilateral ABCD is cyclic ,  m(arc BC) = 90° and angle DBC = 55°  

To find : measure of angle BCD​

Solution:

Let say O is the center of circle passing through quadrilateral vertex

m(arc BC) = 90°

=> ∠BOC = 90°

in Δ BOC

OB = OC = Radius

=> ∠OBC = ∠OCB

∠OBC + ∠OCB + ∠BOC = 180°   ( Sum of angles of triangle)

=> ∠OBC = ∠OCB  = 45°

∠DBC = 55°

=>  ∠DOC = 2∠DBC  = 2 x 55°  = 110°

in Δ DOC

OC = OD = Radius

=> ∠OCD = ∠ODC  = 35°    ( (180 - 110)/2 )

∠BCD =   ∠OCB +  ∠OCD

= 45° + 35°

= 80°

∠BCD =  80°

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