In given figure, BAC = 60° and AB is a diameter, then CBD is:
Answers
Answer:
It is given that AB is a diameter of a circle with centre O and DO || CB
(i) We know that ABCD is a cyclic quadrilateral
It can be written as
∠BCD + ∠BAD = 180o
By substituting the values
120o + ∠BAD = 180o
On further calculation
∠BAD = 180o – 120o
By subtraction
∠BAD = 60o
(ii) We know that the angle in a semi-circle is right angle
∠BDA = 90o
Consider △ ABD
By using the angle sum property
∠BDA + ∠BAD + ∠ABD = 180o
By substituting the values
90o + 60o + ∠ABD = 180o
On further calculation
∠ABD = 180o – 90o – 60o
By subtraction
∠ABD = 180o – 150o
So we get
∠ABD = 30o
(iii) We know that OD = OA
So we get ∠ODA = ∠OAD = ∠BAD = 60o
From the figure we know that
∠ODB + ∠ODA = 90o
By substituting the values
∠ODB + 60o = 90o
On further calculation
∠ODB = 90o – 60o
By subtraction
∠ODB = 30o
It is given that DO || CB
We know that the alternate angles are equal
∠CDB = ∠ODB = 30o
(iv) From the figure we know that
∠ADC = ∠ADB + ∠CDB
By substituting the values
∠ADC = 90o + 30o
By addition
∠ADC = 120o
Consider △ AOD
By using the angle sum property
∠ODA + ∠OAD + ∠AOD = 180o
By substituting the values
60o + 60o + ∠AOD = 180o
On further calculation
∠AOD = 180o – 60o – 60o
By subtraction
∠AOD = 180o – 120o
So we get
∠AOD = 60o
We know that all the angles of the △ AOD is 60o
Therefore, it is proved that △ AOD is an equilateral triangle.
Answer:
answer is "b"
because
BAC=60
ACB=90 (right angil is 90,AB is diameter)
ACB+BAC-CBD=180 (some of triangle angils =180)
90+60-CBD=180
150-180=CBD
CBD=30