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CBSE Questions for Class 9 Maths Circles Quiz 6 - MCQExams.com
CBSE
Class 9 Maths
Circles
Quiz 6
In the given figure, if
O
is centre, then external
∠
A
O
C
=
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0%
190
∘
0%
160
∘
0%
200
∘
0%
185
∘
Explanation
In\quad the\quad given\quad figure\\ \angle ADC+\angle CDE=180^° (Linear\quad pair\quad of\quad angles)\\ \angle CDE=80^° (Given)\\ \angle ADC=180^°-\angle CDE\\ \quad \quad \quad \quad \quad =180^°-80^°=100^°\\ Also,\angle AOC=2\angle ADC\\ [\because Angle\quad at\quad centre\quad is\quad double\quad the\quad angle\quad on\quad remaining\quad part\quad of\quad circle]\\ Hence\angle AOC=2\times 100^°=200^°\\ Hence\quad option(C)\quad is\quad right\quad answer
In the given figure,
\angle DBC=22^{\circ}\;and\;\angle DCB=78^{\circ}
then
\angle BAC
is equal to:
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0%
90^{\circ}
0%
80^{\circ}
0%
78^{\circ}
0%
22^{\circ}
Explanation
Let
\angle BAC = x
\therefore \angle BDC = x
(angle subtended by chord on same segment are equal
)
In
\Delta BDC
,
x+22^{\circ}+78^{\circ} = 180^{\circ}
x+100 = 180^{\circ}
\boxed {x = 80^{\circ}} = \angle BAC
Ans (B)
In the given figure determine the value of x.
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0%
90^{\circ}
0%
115^{\circ}
0%
130^{\circ}
0%
65^{\circ}
Explanation
In the given figure,
CDEF
is a cyclic quadrilateral,
DE\;
and
\;CF
are produced to
A\;
and
\;B
respectively such that
AB\;\parallel\;CD
. If
\angle FED=80^{\circ}
, find
\angle FBA
.
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0%
30^{\circ}
0%
60^{\circ}
0%
80^{\circ}
0%
10^{\circ}
Explanation
CDEF\quad is\quad a\quad cyclic\quad quadrilateral.\\ \angle FED=80^o(Given)\\ \angle FED+\angle FCD=180^o ...(sum\quad of opposite\quad angles\quad of\quad cyclic\quad quadrilateral\quad is\quad 180^o)\\ \Rightarrow FCD=180^o-80^o=100^o.\\ Since\quad AB\parallel CD,\\ \Rightarrow ABCD\quad is\quad a\quad parallelogram.\\ \angle ABC+\angle BCD=180^o ...(Adjacent\quad angles\quad of\quad parallelogram.\quad are\quad supplementary)\\ \angle BCD=\angle FCD=100^o\\ \Rightarrow \angle ABC=180^o-100^o=80^o\\ \angle ABC=\angle FBA=80^o.\\ Hence,\quad option\quad C \quad is\quad correct.
In figure,
O
is centre, then
\angle BXD=
Report Question
0%
65^{\circ}
0%
60^{\circ}
0%
70^{\circ}
0%
55^{\circ}
Explanation
Given
\angle AOC={ 95 }^{ o }
\angle ABC=\angle ADC=\cfrac { 95^o }{ 2 } \; \; (Half\; Angle)\\ \angle EBX=\angle EOX={ 180 }^{ o }-\cfrac { 95^o }{ 2 } =\cfrac { 265^o }{ 2 }
In quadrilateral
BEXD
\angle BED+\angle EBX+\angle BXD+\angle XDE=360^o \\ 25^o+\cfrac{265^o}{2}+\angle BXD+\cfrac{265^o}{2}=360^o \\ \angle BXD=360^o-265^o-25^o \\ \angle BXD=70^o
In the figure,
O
is the center. If
\angle MON=80^o
, then
\angle MQN
equals
Report Question
0%
40^o
0%
160^o
0%
100^o
0%
10^o
Explanation
Angle subtended by an arc at the center is double the angle subtended by the same are at any point on the circumference.
\angle MON\quad =80^o\\ \angle MQN=\frac { 80^o }{ 2 } =40^o
PQ
is a diameter and
PQRS
is a cyclic quadrilateral. If
\angle PSR=150^o
, then measure of
\angle RPQ
is:
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0%
90^{\circ}
0%
60^{\circ}
0%
30^{\circ}
0%
None of these
Explanation
PQRS
is a cyclic quadrilateral.
Then,
\angle PQR +\angle PSR= 180^{\circ}
...[
Opposite angles of cyclic quadrilateral are supplementary
]
\implies
\angle PQR +150^o= 180^{\circ}
\implies
\angle PQR = 180^{\circ} - 150^{\circ} = 30^{\circ}
.
In
\triangle PQR
,
\angle PRQ = 90^{\circ}
(Angle of a semicircle)
Then,
\angle RPQ + 90^{\circ} + 30^{\circ} = 180^{\circ}
...[Angle sum property]
\Rightarrow \angle RPQ + 120^{\circ} = 180^{\circ}
\Rightarrow \angle RPQ = 60^{\circ}
.
Hence, option
B
is correct.
ABCD
is a cyclic quadrilateral inscribed in a circle with the centre
O
. Then
\angle OAD
is equal to:
Report Question
0%
30^{\circ}
0%
40^{\circ}
0%
50^{\circ}
0%
60^{\circ}
Explanation
Given-\quad \\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral\quad inscribed\quad in\quad a\\ circle\quad with\quad centre\quad O.\\ OA,\quad OB,\quad OC\quad \& \quad OD\quad have\quad been\quad joined.\\ \angle OAB={ 40 }^{ o },\quad \angle OBC={ 30 }^{ o }\quad \& \quad \angle OCD={ 50 }^{ o }.\\ To\quad find\quad out-\\ \angle OAD=?\\ Solution-\\ OC=OB\quad (radii\quad of\quad the\quad same\quad circle)\\ \therefore \quad \Delta OBC\quad is\quad an\quad isosceles\quad triangle.\\ \therefore \quad \angle OCB=\angle OBC={ 30 }^{ o }.\\ \therefore \quad \angle BCD=\angle OCD+\angle OCB={ 50 }^{ o }+{ 30 }^{ o }={ 80 }^{ o }.\\ Now,\quad ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ So,\quad by\quad angle\quad sum\quad property\quad of\quad a\quad cyclic\quad quadrilateral,\\ we\quad get,\\ \angle BAD={ 180 }^{ o }-\angle BCD={ 180 }^{ o }-{ 80 }^{ o }={ 100 }^{ o }.\\ So,\quad \angle OAD=\angle BAD-\angle OAB={ 100 }^{ o }-{ 40 }^{ o }={ 60 }^{ o }.\\ Hence, \quad option\quad D \quad is \quad correct.
ABCD
is a cyclic quadrilateral. Then, find
\angle x^{\circ}
as given in the figure.
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0%
50^{\circ}
0%
80^{\circ}
0%
90^{\circ}
0%
100^{\circ}
Explanation
\angle EDC+\angle ADC=180^o ...(Linear\quad pair\quad of\quad angles)\\ \angle EDC=80^o ...(Given)\\ Then, \angle EDC+\angle ADC=180^o\\ \angle ADC=180^o-\angle EDC=180^o-80^o=100^o.\\ ABCD\quad is\quad a\quad cyclic\quad quadrilateral.\\ Hence,\quad \angle ADC+\angle ABC=180^o ...(Sum\quad of\quad opposite\quad angles\quad is\quad 180^o)\\ \Rightarrow \angle ABC=180^o-\angle ADC\\ \quad \quad \quad \quad \quad \quad \quad =180^o-100^o\\ \quad \quad \quad \quad \quad \quad \quad =80^o.\\ Now, \angle ABF+\angle ABC=180^o\\ \angle ABF=180^o-\angle ABC\\ \angle ABF=\angle x ...(Given)\\ \Rightarrow \angle x=180^o-80^o\\ \Rightarrow \angle x=100^o.\\ Hence,\quad option \quad D \quad is\quad correct.
In the given figure, the value of
'a'
is:
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0%
30^{\circ}
0%
40^{\circ}
0%
60^{\circ}
0%
90^{\circ}
Explanation
Given-
\overline { AOB }
is a diameter of a given circle.
ABCD
is a cyclic quadrilateral.
AC
is joined. Also,
\angle ADC={ 130 }^{ \circ }
.
Since,
\overline { AOB }
is the diameter of the given circle, it subtends
\angle ACB
to the circumference at
C
, i.e.
\angle ACB={ 90 }^{ \circ }
...[ since it is an angle in a semicircle].
Again,
\angle ADC+\angle ABC=180^\circ
...[ sum of opposite angles of a cyclic quadrilateral is
{ 180 }^{ \circ }
]
\Rightarrow \angle ABC={ 180 }^{ \circ }-\angle ADC
\Rightarrow \angle ABC={ 180 }^{ \circ }-130^{ \circ }
\Rightarrow\angle ABC={ 42 }^{ \circ}
.
In
\triangle ABC,
\angle ABC +\angle ACB+\angle CAB=180^\circ
\angle CAB={ 180 }^{ \circ }-(\angle ACB+\angle ABC)
={ 180 }^{ \circ }-({ 90 }^{ \circ }+{ 50 }^{ \circ })
={ 40 }^{ \circ }
Hence, option
B
is correct.
In the circle, reflex
\angle AOC=x^o, \angle ABC=y^o
.
OABC
is a parallelogram, then find
y
.
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0%
80^o
0%
120^o
0%
110^o
0%
Data insufficient
Explanation
In the circle, clearly,
x=2y
.
Then,
\angle A OC=360-x=360-2y
.
Here,
ABCD
is a parallelogram
Then,
{BCO}+ {AOC}=180^o
\Rightarrow {BCO}+360-2y=180^o
\Rightarrow 2y=BCO+180^o
.
Similarly,
ABC+BCO=180^o
\Rightarrow y+2y-180=180
\Rightarrow 3y=360^o
\Rightarrow y=120^o
.
Hence, option
B
is correct.
Find
\angle ASR
.
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0%
52^{\circ}
0%
78^{\circ}
0%
102^{\circ}
0%
10^{\circ}
Explanation
PABQ\quad is\quad a\quad cyclic\quad quadilateral\\ \angle QPA=78^o...(Given). \\ Then, \angle QPA+\angle ABQ=180^o\\ \Rightarrow 78^o+\angle ABQ=180^o\\ \Rightarrow \angle ABQ=180^o-78^o=102^o....(1).\\ \angle ABR\quad and\quad \angle ABQ\quad are\quad linear\quad pair\quad of\quad angles\quad on\quad straight\quad line\quad QR\\ Then, \angle ABQ+\angle ABR=180^o\\ \Rightarrow \angle ABR=180^o-\angle ABQ=180^o-102^o=78^o.\\ Hence,\quad \angle ABR\quad =78^o.\\ In\quad cyclic\quad quadilateral\quad ABCD,\quad \\ \angle ASR+\angle ABR=180^o\\ \Rightarrow \angle ASR=180^o-78^o\\ \Rightarrow \angle ASR=102^o.\\ Hence,\quad option\quad C \quad is\quad correct.
In the given figure,
PQRS
is a cyclic quadrilateral. Its diagonals
PR
and
QS
intersect each other at
T
. If
\angle PRS=80^{\circ}\;and\;\angle RQS=50^{\circ}
, calculate
\angle PSR
.
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0%
30^{\circ}
0%
50^{\circ}
0%
70^{\circ}
0%
90^{\circ}
Explanation
Given:\\ PQRS\ is\ a\ cyclic\ quadilateral.\\ \angle PRS=80^o....(1)\\ \angle RQS=50^o\\ \angle RQS=\angle RPS\ (Angles\ in\ the\ same\ segment\ are\ equal)\\ \Rightarrow \angle RPS=50^o...(2)\\ In\quad \triangle PSR\\ \angle PRS+\angle RPS+\angle PSR=180^o\ (Angle\ sum\ property)\\ 80^o+50^o+\angle PSR=180^o\ (From\ (1)\ and\ (2))\\ \Rightarrow \angle PSR=180^o-130^o\\ \ \ \ \ \angle PSR=50^o\\ Hence\ option\ (B)\ is\ right\
In figure,
\angle BAC = 60^{\circ}
and
\angle BCA = 20^{\circ}
, find
\angle ADC
.
Report Question
0%
15
^{\circ}
0%
50
^{\circ}
0%
80
^{\circ}
0%
40
^{\circ}
Explanation
In
\Delta ABC
, we have
\angle BAC + \angle BCA + \angle ABC = 180^{\circ}
....[Angle sum property]
\Rightarrow 60^{\circ} + 20^{\circ} + \angle ABC = 180^o
\therefore \angle ABC = 180^{\circ} - 80^{\circ} = 100^{\circ}
.
ABCD
is a cyclic quadrilateral,
\therefore \angle ABC + \angle ADC = 180^{\circ}
.....
[Since, opposite angles of a cyclic quadrilateral are supplementary]
\implies
100^{\circ} + \angle ADC = 180^{\circ}
\implies \angle ADC = 180^{\circ} - 100^{\circ} = 80^{\circ}
.
Hence, option
C
is correct.
In the figure,
\angle
B is equal to:
Report Question
0%
85^0
0%
95^0
0%
70^0
0%
115^0
Explanation
Since,
APQD
is a cyclic quadrilateral,
therefore
\angle A + \angle DQP = 180^o
...(opposite angles of cyclic quadrilateral are supplementary)
\implies
\angle DQP = 180^o -85^o =95^o
.
Now,
\angle PQC = 180^o - \angle DQP=180^o-95^o=85^o
...[linear pairs].
\therefore \angle B = 180^o -85^o =95^o
...(opposite angles of cyclic quadrilateral are supplementary).
Hence, option
B
is correct.
In the figure above, if O is the centre of the circle, find the value of y.
Report Question
0%
40
^{\circ}
0%
70
^{\circ}
0%
60
^{\circ}
0%
30
^{\circ}
Explanation
Given that, the centre of the circle is
O
and
\angle BOD=140^\circ
To find out: The value of
y
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circle.
\therefore \ \angle BOD = 2 \angle BAD
\Rightarrow 140^{\circ} = 2 \angle BAD
\displaystyle \Rightarrow \angle BAD = \frac{1}{2} \times 140^{\circ} = 70^{\circ}
Now, in cyclic quadrilateral ABCD,
\angle BAD + \angle BCD = 180^{\circ}
[Opposite angles of a cyclic quadrilateral are supplementary]
\therefore \ 70^{\circ} + \angle BCD = 180^{\circ}
\therefore \angle BCD = 180^{\circ} - 70^{\circ} = 110^{\circ}
Also,
\angle BCD + \angle DCP = 180^{\circ}
[Linear pair]
\Rightarrow 110^{\circ} + y = 180^{\circ}
\therefore \ y = 70^{\circ}
Hence, option B is correct.
In a cyclic quadrilateral
ABDC,\,\angle CAB=80^{\circ}
and
\angle ABC=40^{\circ}
. The measure of the
\angle ADB
will be?
Report Question
0%
120^{\circ}
0%
80^{\circ}
0%
60^{\circ}
0%
40^{\circ}
Explanation
In
\Delta ACB
,
\because\;\angle ACB=180^{\circ}-(80^{\circ}+40^{\circ})=60^{\circ}
\therefore\;\angle ADB=\angle ACB
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=60^{\circ}
In fig, a is the centre of the circle. If
\angle
BOD 160
^{\circ}
find the values of x and y.
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0%
80
^{\circ}
, 100
^{\circ}
0%
135
^{\circ}
, 45
^{\circ}
0%
140
^{\circ}
, 40
^{\circ}
0%
120
^{\circ}
, 60
^{\circ}
Explanation
In the cyclic quadrilateral ABCD
\displaystyle \angle BCD = \frac{1}{2} \angle BOD
[Angle made by an arc at the centre is double the angle made by it, at any other point on the remaining part of the circle]
\displaystyle \therefore \angle x = \frac{1}{2} \times 160^{\circ} = 80^{\circ}
\therefore x = 80^{\circ}
\angle x + \angle y = 180^{\circ}
[opp. angles of a cyclic quadrilateral]
\therefore \angle y = 180^{\circ} - \angle x
\angle y = 180^{\circ} - 80^{\circ} = 100^{\circ}
y = 100^{\circ}
One of the base angle a cyclic trapezium is double the other. What is the measure of the larger angle?
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0%
60^0
0%
80^0
0%
75^0
0%
120^0
Explanation
In cyclic trapezium,
Sum of opposite angle
=180^o
...[opposite angles of a cyclic quadrilateral are supplementary].
Let the smaller angle be
x
and larger angle
=2x
.
Then,
2x+x=180^o \implies 3x=180^o \implies x=60^o
.
Therefore, Larger angle
=2x=2\times 60^o=120^o
.
Hence, option
D
is correct.
Find
x.
Report Question
0%
40^{\circ}
0%
50^{\circ}
0%
30^{\circ}
0%
60^{\circ}
Explanation
ABCD is a quadrilateral inside the circle,
\angle D + \angle B = 180^o
130^o + \angle B = 180^o
\angle B = 50^o
AB is diameter
\angle C = 90^o
[angle in a semicircle]
In
\Delta ABC
\angle A + \angle B + \angle C = 180^o
\angle A + 50^o + 90^o = 180^o
\Rightarrow \angle A + 140^o = 180^o
\Rightarrow \angle A = 180^o - 140^o
=40^o
\therefore \angle CAB = 40^o \Rightarrow x = 40^o
option (A) is correct.
ABCD is cyclic quadrilateral. The tangents to a circle at A and C meet at P. If
\angle
APC = 50
^{\circ}
, then the value of
\angle
ADC is:
Report Question
0%
\displaystyle 65^{\circ}
0%
\displaystyle 70^{\circ}
0%
\displaystyle 80^{\circ}
0%
none of these
Explanation
Given
\angle APC=50^0
As the radius is always perpendicular to tangent at the point of contact,
\angle OCP=\angle OAP=90^0
As
AOCP
is a quadrilateral, sum of all its interior angles are equal to
360^0
\therefore\ \angle AOC=130^0\quad \quad [360^0-(90^0+90^0+50^0)]
We know that, angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc on the circumference of the circle.
\therefore \ \angle ADC=\dfrac{130^0}{2}
=65^0
Hence,
\angle ADC=65^0
.
If
O
is the centre of the circle and
A, B
and
C
are points on its circumference and
\angle AOC = 130^{\circ}
, find
\angle ABC
.
Report Question
0%
105^{\circ}
0%
115^{\circ}
0%
110^{\circ}
0%
120^{\circ}
Explanation
Take any point
P
on the circumference of the circle as shown.
Join
AP
and
CP
.
\because ABC
subtends
\angle AOC
at centre
O
and
\angle APC
at any point
P
on the circumference of the circle.
\therefore \angle AOC = 2 \angle APC
[
\because
Angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circumference]
\displaystyle \implies \angle APC = \frac{1}{2} \angle AOC ~~~[ \because AOC = 130^{\circ}]
\implies
\displaystyle \frac{1}{2} \times 130^{\circ} = 65^{\circ}
.
\because
ABCP
is a cycle quadrilateral,
\implies \angle APC + \angle ABC = 180^{\circ}
...
[
\because
sum of opposite angles of a cyclic quadrilateral is
180^{\circ}
]
\implies 65^{\circ} + \angle ABC = 180^{\circ}
\therefore \angle ABC = 180^{\circ} - 65^{\circ} = 115^{\circ}
.
Hence, option
B
is correct.
ABCD
is a cyclic quadrilateral, in which
BC
is parallel to
AD
,
\displaystyle\angle ADC=\displaystyle 110^{\circ}
and
\displaystyle \angle BAC=50^{\circ}
. What is the value of
\displaystyle \angle DAC
?
Report Question
0%
\displaystyle 40^{\circ}
0%
\displaystyle 50^{\circ}
0%
\displaystyle 60^{\circ}
0%
\displaystyle 64^{\circ}
Explanation
ABCD
is a cyclic quadrilateral.
Then,
\angle ABC+\angle ADC=180^o
...[Opposite angles of a cyclic quadrilateral are supplementary]
\Rightarrow
\angle ABC =180^{\circ}-110^{\circ} =70^{\circ}
.
In
\triangle ABC
,
\angle ABC+\angle ACB+\angle BAC=180^o
...[Angle sum property]
\Rightarrow
70^o+\angle ACB+50^o=180^o
\Rightarrow
\angle ACB=180^{\circ}-(50^{\circ}+70^{\circ})=60^{\circ}
.
Given,
BC
and
AD
are parallel,
\Rightarrow \angle ACB=\angle DAC =60^{\circ}
...[Alternate interior angles].
Hence, option
C
is correct.
In the given figure, AB=AC and
\displaystyle \angle ABC=68^{\circ},
what is the value of
\displaystyle \angle BPC
?
Report Question
0%
\displaystyle 40^{\circ}
0%
\displaystyle 42^{\circ}
0%
\displaystyle 44^{\circ}
0%
\displaystyle 48^{\circ}
Explanation
Theorem: Chord of a circle subtends equal angles at different points on the circumference of the circle on same side
\Rightarrow \angle BAP=\angle BPC
AB=AC\Rightarrow \angle B=\angle C=68^{°}
(Opposite angles of equal sides are equal)
In
\triangle ABC
\angle A+ \angle B + \angle C =180^o
(Sum of angles in a triangle )
68^o+68^o+ \angle A = 180^o
136^o+\angle A = 180^o
\Rightarrow \angle A=44^{°}
\Rightarrow \angle A=\angle BAC =\angle BPC =44^{°}
(Angles in same segment are equal )
In the given figure,
BD=DC
and
\displaystyle \angle DBC=30^{\circ}
what is the value of
\displaystyle \angle BAC
?
Report Question
0%
\displaystyle 40^{\circ}
0%
\displaystyle 50^{\circ}
0%
\displaystyle 60^{\circ}
0%
\displaystyle 65^{\circ}
Explanation
As
BD=DC
......
(given)
\Rightarrow \angle DBC=\angle DCB=30^{\circ}
\Rightarrow \angle BDC =180^\circ -30^\circ-30^\circ=120^{\circ}
As
ABCD
is a cyclic quadrilateral ;
\angle BAC+\angle BDC =180^{\circ}
\Rightarrow \angle BAC =180^{\circ}-120^{\circ} =60^{\circ}
What is the value of
\displaystyle \angle ABC
?
Report Question
0%
\displaystyle 105^{o}
0%
\displaystyle 110^{o}
0%
\displaystyle 115^{o}
0%
\displaystyle 120^{o}
Explanation
Extending A and C onto the circumference to D as shown in the figure
Theorem: Angle subtended by an arc at the centre is double the angle subtended by the same arc at the circumference of a circle
\Rightarrow \angle AOC=2\times \angle ADC
\Rightarrow \angle ADC=\dfrac {130}{2}=65^o
As
ABCD
is a cyclic quadrilateral, sum of its opposite interior angles are equal to
180^o
\Rightarrow \angle ADC+\angle CBA =180^o
\Rightarrow \angle ABC =180^o - 65^o=115^o
ABCD is a cyclic trapezium and
\displaystyle AB\parallel CD
. If
AB
is the diameter of the circle and
\displaystyle \angle CAB=30^{\circ}
, then the value of
\displaystyle \angle ADC
is:
Report Question
0%
\displaystyle 110^{\circ}
0%
\displaystyle 115^{\circ}
0%
\displaystyle 120^{\circ}
0%
\displaystyle 125^{\circ}
Explanation
Here,
\angle CAB=30^o
...[Given].
Also,
\angle ACB=90^o
...[Angle inscribed in a semi-circle].
Now, in
\triangle ACB
,
\angle ACB+\angle CAB+\angle ABC=180^o
...[Angle sum property]
\Rightarrow
90^o+30^o+\angle ABC=180^o
\Rightarrow
\angle ABC=180^o-120^o
\Rightarrow
\angle ABC=60^o
.
Here,
ABCD
is a cyclic quadrilateral.
We know that, sum of opposite angles of a cyclic quadrilateral is
180^o
.
\therefore
\angle ABC+\angle ADC=180^o
\Rightarrow
60^o+\angle ADC=180^o
\Rightarrow
\angle ADC=180^o-60^o
\Rightarrow
\angle ADC=120^o.
Hence, option
D
is correct.
Find the value of x from the given figure in which O is the center of the circle
Report Question
0%
\displaystyle x=50^{\circ}
0%
\displaystyle x=60^{\circ}
0%
\displaystyle x=80^{\circ}
0%
\displaystyle x=90^{\circ}
Explanation
\textbf{Step - 1 : Find the angle at the center,}
\text{Given that, angle subtend by segment BC at point A on circle}=40^{\circ}
\text{Then, the angle subtend by segment BC at center O}=\angle{\text{BOC}}=40^{\circ}\times2=80^{\circ}
\textbf{Step - 2 : Find the value of x}
\text{In triangle OBC, OB = OC, as the radii of the same circle.}
\text{Hence, }\angle\text{OBC}=\angle\text{OCB}=x, \text{as the angles opposite to equal sides are also equal}
\text{Then by angle sum property of triangle,}
\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^{\circ}\Rightarrow80^{\circ}+x+x=180^{\circ}\Rightarrow2x=180^\circ-80^\circ=100^\circ
\therefore{x}=\dfrac{100^\circ}{2}=50^\circ
\textbf{Therefore, option A }\mathbf{x=50^\circ}\textbf{ is correct answer.}
Find the values of
y
and
z
, where
O
denotes the centre of the circle.
Report Question
0%
y=70^o
and
z=35^o
0%
y=290^o
and
z=145^o
0%
y=145^o
and
z=290^o
0%
y=35^o
and
z=70^o
Explanation
Since the given quadrilateral is a cyclic quadrilateral,
\angle y + 35^o =180^o
\Rightarrow \angle y=180^o-35^o
\Rightarrow \angle y=145^o
.
By the inscribed angle theorem, we have,
\angle z=2\times \angle y
\Rightarrow \angle z=2\times145^o
\Rightarrow \angle z = 290^o
.
Hence, option
C
is correct.
In given figure, BC is a diameter of the circle and
\displaystyle \angle BAO=60^{\circ}\,
. Then
\,\angle ADC
is equal to
Report Question
0%
\displaystyle 30^{\circ}
0%
\displaystyle 45^{\circ}
0%
\displaystyle 60^{\circ}
0%
\displaystyle 120^{\circ}
Explanation
In
\triangle AOB
,
OA=OB
=>\angle OAB=\angle OBA=60^{0}
We know that the exterior angle of a triangle is equal to sum of the two opposite angles of the triangle.
\angle AOC=\angle OAB+\angle OBA
\angle AOC=60^{0}+60^{0}
\angle AOC = 120^{0}
We know that the angle subtended at the centre of the circle by an arch is twice the angle subtended at the circumference by the same arc.
\therefore \angle ADC=\dfrac{1}{2}\angle AOC
=\dfrac{120}{2}
=60^{0}
0:0:2
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