RD Chapter 16 Circles Ex 16.5 Solutions
Question - 1 : - In the given figure, ΔABC is an equilateral triangle. Find m∠BEC.
Answer - 1 : -
It is given that, 
is an equilateral triangle We have to find 
Since is an equilateral triangle. So 
And
…… (1)Since, quadrilateral BACE is a cyclic qualdrilateral
So , 
(Sum of opposite angles of cyclic quadrilateral is .) Hence
Answer - 2 : -
Disclaimer: Figure given in the book was showing m∠PQR as m∠SQR.
It is given that ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°
We have to find the m∠QSR and m∠QTR
Since ΔPQR is an isosceles triangle
So ∠PQR = ∠PRQ = 35°
Then
Since PQTR is a cyclic quadrilateral
So
In cyclic quadrilateral QSRT we have
Hence,
and 
Question - 3 : - In the given figure, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.
Answer - 3 : -
It is given that O is centre of the circle and ∠BOD = 160°
We have to find the values of x and y.
As we know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,
Since, quadrilateral ABCD is a cyclic quadrilateral.
So,
x + y = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°.)
Hence 
and 
Question - 4 : - In the given figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.
Answer - 4 : -
It is given that ∠BCD = 100° and ∠ABD = 70°
We have to find the ∠ADB
We have
∠A + ∠C = 180° (Opposite pair of angle of cyclic quadrilateral)
So,
Therefore,
Hence, 
Answer - 5 : -
It is given that, ABCD is cyclic quadrilateral in which AD || BC
We have to prove 
Since, ABCD is a cyclic quadrilateral
So,
and 
..… (1)
and 
(Sum of pair of consecutive interior angles is 180°) …… (2)From equation (1) and (2) we have
…… (3) 
…… (4) Hence Proved
Question - 6 : - In the given figure, O is the centre of the circle. Find ∠CBD.
Answer - 6 : -
It is given that, 
We have to find 
Since, (Given) So,
(The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.)
Now,
(Opposite pair of angle of cyclic quadrilateral) So,
…… (1) 
(Linear pair) 
(
) Hence 
Question - 7 : - In the given figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 50°, find ∠AOC.
Answer - 7 : -
It is given that, AB and CD are diameter with center O and
We have to find 
Construction: Join the point A and D to form line AD
Clearly arc AD subtends 
at B and 
at the centre. Therefore, ∠AOD=2∠ABD=100°∠AOD=2∠ABD=100° …… (1)
Since CD is a straight line then
∠DOA+∠AOC=180° (Linear pair)∠DOA+∠AOC=180° Linear pair
Hence 
Question - 8 : - On a semi-circle with AB as diameter, a point C is taken, so that m (∠CAB) = 30°. Find m (∠ACB) and m (∠ABC).
Answer - 8 : -
It is given that, 
as diameter, 
is centre and 
We have to find 
and 
Since angle in a semi-circle is a right angle therefore
In 
we have (Given) 
(Angle in semi-circle is right angle) Now in 
we have Hence 
and 
Question - 9 : - In a cyclic quadrilateral ABCD if AB || CD and ∠B = 70°, find the remaining angles.
Answer - 9 : -
It is given that, ABCD is a cyclic quadrilateral such that AB || CD and 
Sum of pair of opposite angles of cyclic quadrilateral is 180°.
( given) So, 
Also AB || CD and BC transversal
So,
Now
Question - 10 : - In a cyclic quadrilateral ABCD, if m ∠A = 3 (m ∠C). Find m ∠A.
Answer - 10 : -
It is given that
ABCD is cyclic quadrilateral and 
We have to find 
Since ABCD is cyclic quadrilateral and sum of opposite pair of cyclic quadrilateral is 180°.
So And
Therefore
Hence 