Maths M.C.Q

 $\triangle\text{ABC}$ is an equilateral triangle so $\angle\text{BAC}=60^\circ$
In cyclic quadrilateral $ABCD,$ we have:
$\angle\text{BDC}+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BDC}+60^\circ=180^\circ$
$\therefore\angle\text{BDC}=(180^\circ-60^\circ)=120^\circ$

We know that the line joining their centres is the perpendicular bisector of the common chord.
Join $AP.$
Then $A P=5 \mathrm{~cm} ; A B=4 \mathrm{~cm}$
Also, $A P^2=B P^2+A B^2$ [using pythagoras theorem]
$\Rightarrow B P^2=A P^2-A B^2$
$\Rightarrow B P^2=5^2-4^2$
$\Rightarrow B P=3 \mathrm{~cm}$
$\therefore$ triangle $A B P$ is a right angled and $P Q=2 \times B P=(2 \times 3) \mathrm{cm}=6 \mathrm{~cm}$

$OB = BC [$Given$]$


$\Rightarrow\ \angle\text{OBC}=\angle\text{BCO}=25^\circ$ [Angles opposite equal sides are equal]
Now,
$\angle\text{OBC}=\angle\text{BOC}+\angle\text{BCO}=25^\circ+25^\circ=50^\circ$
$OA = OB [$Radii of the same circle$]$
$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ$
In $\triangle\text{AOC},$
$\angle\text{AOD}=\angle\text{OAC}+\angle\text{ACO}$
$=\angle\text{OAB}+\angle\text{BCO}$
$=50^\circ+25^\circ$
$=75^\circ$

 
Let $L$ and $M$ be the midpoints of $R s$ and $P Q$ respectively.
Let $O M=x$ thus, $O L=17-x$
Now in triangle $RLO, RL = 12$ and
$\mathrm{RL}^2+O L^2=\mathrm{r}^2$
$12^2+(17-\mathrm{x})^2=r^2 .....(i)$
Similarly,
In triangle $O M P, P M=5$ and
$P M^2+O M^2=O P^2$
$(x)^2+5^2=r^2 \ldots . .(\text { ii) }$
Equating $(i)$ and $(ii),$ we get :-
$12^2+(17-x)^2=x^2+5^2$
$\left.144+289-34 x+x^2=x^2+25 \text { \{using, }(a-b)^2\right\}$
On solving, we get:
$\mathrm{x}=\frac{408}{34}=12$
So, $17-x=17-12=5$
Thus, from$(i),$
$R L^2+O L^2=r^2$
$12^2+5^2=r^2$
$144+25=r^2$
$r^2=169$
so, radius $=\sqrt{169}$
Hence, the radius is $13\ cm.$

We are given the major arc is $3$ times the minor arc. We are asked to find the corresponding central angle.
See the corresponding figure.

We know that the angle formed by the circumference at the centre is $360^\circ .$
Since the circumference of the circle is divided into two parts such that the angle formed by major and minor arcs at the centre are $3x$ and $x$ respectively.
So $3x + x = 360$
$4x = 360$
$x = 90$
So $m\ AB = 90^\circ $ and $m\ AB = 3x = 270^\circ $
the degree measures of two arcs are $90^\circ$ and $270^\circ$

We need to find a common chord.
We have the corresponding figure as follows:

$AO = AO' = r ($radius$)$
And $OO' = r$
So, $\triangle\text{OAO}'$ is an equilateral triangle.
We know that the attitude of an equilateral triangle with side $r$ is given by $\frac{2}{\sqrt{3}}$
That is $\text{AM}=\frac{\sqrt{3}}{2\text{r}}$
We know that the line joining centre of the circles divides the common chord into two equal parts.
So we have
$\text{AB}=2\text{AM}=2.\frac{\sqrt{3}}{2\text{r}}$
$\text{AB}=\sqrt{3}\text{r}$

Angle made by a chord at the centre is twice the angle made by it on any point of the circumference.
So, $\angle\text{AOC}=2\angle\text{ABC}=2\times20^\circ=40^\circ$

 
$\text{m}\widehat{\text{AB}}=183^\circ$
$O$ is the center of the circle and $AB$ is a chord.
The region bounded by arc and line segment $AB$ is shaded.
We can see, $'O',$ the center, always lie in the interior of $S.$

 Here, given
$OP = OQ$ and $OR = OQ ($Radius of circle$)$
So, {angles opposite to equal sides are also equal}
Hence,
$PQR = 25^\circ + 20^\circ = 45^\circ $
and $PQR = 2 PQR = 2 45^\circ = 90^\circ $
{Angle subtended by same sides on centre is double the angle at opposite vertex}

To solve this problem we need to know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

Here we are given that $A, B, C,$ are three points on a circle with centre $O$ such that
$\angle\text{AOB}=90^\circ$ and $\angle\text{BOC}=120^\circ.$
From
$\angle\text{AOC}=360^\circ-\angle\text{AOB}-\angle\text{BOC}$
$\Rightarrow\angle\text{AOC}=360^\circ-90^\circ-120^\circ$
$\Rightarrow\angle\text{AOC}=360^\circ-210^\circ$
$\Rightarrow\angle\text{AOC}=150^\circ$
Now, as seen earlier, the angle made by the arc $AC$ with the centre of the circle will be twice the angle it makes in any point in the remaining part of the circle.
Since the point $C$ lies on the remaining part of the circle, the angle the arc $AC$ makes with this point has to be half of the angle $'AC'$ makes with the centre. Therefore we have,
$\angle\text{ABC}=\frac{\angle\text{AOC}}{2}=\frac{150^\circ}{2}=75^\circ$
$\Rightarrow\angle\text{ABC}=75^\circ$

$\angle\text{ACO}=\angle\text{CAO}=20^\circ ($because $OA = OC)$
$\angle\text{OBC}=\angle\text{OCB}=35^\circ ($because $OB = OC)$
$\angle\text{ACB}=55^\circ$
$\text{x}=2\angle\text{ACB}=2\times55^\circ=110^\circ$

We have:
$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=(\frac{1}{2}\times90^\circ)=45^\circ$
$\Rightarrow\angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=(2\times30^\circ)=60^\circ$
$\therefore\angle\text{COD}=180^\circ-\angle\text{COA}=(180^\circ-60^\circ)=120^\circ$
$\Rightarrow\angle\text{CAO}=\frac{1}{2}\angle\text{COD}=(\frac{1}{2}\times120^\circ)=60^\circ$
$\Rightarrow\angle\text{CAO}=60^\circ$

Proof: $BA$ is the major arc and angle $ADB$ is the angle formed by it in the alternate segment.
Therefore $2\angle\text{ADB}=\text{m}(\text{arc BA})$
$\Rightarrow2\angle\text{ADB}=360-\text{m}(\text{arc AB})$
$\Rightarrow2\angle\text{ADB}=360-\angle\text{AOB}(\text{since}(\text{arc AB})=\text{angle AOB})$
$\Rightarrow2\angle\text{ADB}>360-180(\text{since}\angle\text{AOB}<180)$
$\Rightarrow\angle\text{ADB}>90^\circ$
$\Rightarrow\angle\text{ADB}$ is obtuse.

Only $1$ circle can be drawn from three non-collinear points.

 Given: $AB = CD$
We know that equal chords of a circle subtend equal angles at the centre.
$\therefore\angle\text{COD}=\angle\text{AOB}=80^\circ$
$\Rightarrow\angle\text{COD}=80^\circ$

In $\triangle\text{AOB}$
$\angle\text{A}=\angle\text{B}=40^\circ$
And $\angle\text{A}+\angle\text{B}+\angle\text{O}=180^\circ$
$\Rightarrow\angle\text{O}=100^\circ$
So, $\angle\text{ACB}=\frac{100^\circ}{2}=50^\circ$

We are given that an equilateral $\triangle\text{ABC}$ is inscribed in a circle with centre $O.$ We need to find $\angle\text{BOC}.$
We have the following corresponding figure.

We are given $AB = BC = AC$
Since the sides, $AB, BC,$ and $AC$ are these equal chords of the circle.
Hence,
$\angle\text{AOB}+\angle\text{BOC}+\angle\text{AOC}=360$
$\Rightarrow\angle\text{BOC}+\angle\text{BOC}+\angle\text{BOC}=360$
$\Rightarrow3\angle\text{BOC}=360$
$\Rightarrow\angle\text{BOC}=\frac{360}{3}$
$\Rightarrow\angle\text{BOC}=120^\circ$

Since, $AB$ is a chord and makes $\angle\text{APB}=70^\circ$ at the circumference, so $\angle\text{AOB}=140^\circ$
Now, as $AOBC$ is a cyclic quadrilateral then, sum of opposite angles must be $180^\circ .$
$\angle\text{AOB}+\angle\text{ACB}=180^\circ$
$\Rightarrow140^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=40^\circ$

$\angle\text{APC}+\angle\text{APB}=180^\circ$
$\Rightarrow\angle\text{APB}=180^\circ-90^\circ=90^\circ$
In $\triangle\text{APB},$
$\angle\text{ABP}=180^\circ-\angle\text{APB} -\angle\text{BAP}$
$180^\circ-90^\circ-35^\circ=55^\circ$
Now Arc $\widehat{\text{AC}}$ makes $\angle\text{ABC}$ and $\angle\text{ADC}$ on circle.
$\Rightarrow\text{ABC}=\angle\text{ADC}$
$\Rightarrow\angle\text{ADC}=55^\circ$

$\angle\text{AOB}=60^\circ$ $($Since $ \triangle\text{AOB}$ is equilateral triangle$)$
Now, $\angle\text{ADB}=30^\circ$
$($Since chord $AB$ makes $60$ at centre, same chord will make half of the angle at circumference of angle made at centre$)$
Now $\angle\text{ACB}$ is angle made by chord at minor arc of circle.
$ABCD$ is cyclic Quadrilateral.
$\Rightarrow\angle\text{C}+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{ACB}+\angle\text{ADB}=180^\circ$
$\Rightarrow\angle\text{ACB}=180^\circ-30^\circ=150^\circ$

Since $BOC$ is a diameter of a circle, $\angle\text{BAC}$ is $90^\circ .$
Given that $AB = AC.$
$\Rightarrow\ \angle\text{ABC}=\angle\text{ACB}$
In $\triangle\text{BAC},$
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ [Angle sum property]
$\Rightarrow\ \angle\text{ABC}+\angle\text{ABC}+90^\circ=180^\circ$
$\Rightarrow\ 2\angle\text{ABC}=90^\circ$
$\Rightarrow\ \angle\text{ABC}=45^\circ$

As we know that equal chords make an equal angle at the center.
Therefore,
$\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}$
$\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}=180^\circ [$Sum of Linear pair of angle is $180^\circ ]$
$\Rightarrow3\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=60^\circ$

$\angle\text{BDC}=\angle\text{BAC}=60^\circ$ (Angles in the same segment of a circle)
In $\triangle\text{BDC},$ we have
$\angle\text{DBC}+\angle\text{BDC}+\angle\text{BCD}=180^\circ$ (Angle sum property of a triangle)
$\therefore50^\circ+60^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-(50^\circ+60^\circ)=(180^\circ-110^\circ)=70^\circ$
$\Rightarrow\angle\text{BCD}=70^\circ$

Since $\angle\text{CDB}=\angle\text{CAB}$
So, $\angle\text{CAB}=40^\circ$
Now $\angle\text{PAC}+\angle\text{CAB}=180^\circ$ [Linear Pair]
Hence, $\angle\text{PAC}=14^\circ$

$\angle\text{CDB}=\frac{110^\circ}{2}=55^\circ$
Now, $\angle\text{ADB}=90^\circ$ (Angle in a semicircle)
So, $\angle\text{ADC}=\text{x}=90^\circ-55^\circ=35^\circ$

Construction: Join $OD.$

$O B=O D$
$\Rightarrow O B=O E+E B$
$\Rightarrow O B=O E+4$
$\Rightarrow O E=O B-4 \ldots \text { (i) }$
$\text { In right } \triangle O E D$
$O D^2=O E^2+D E^2$
$\Rightarrow O D^2=(O B-4)^2+D E^2[\text { From (i) }]$
$\Rightarrow O D^2=O B^2-80 A+16+8^2$
$\Rightarrow 8 O D=16+64$
$\Rightarrow 8 O D=80$
$\Rightarrow O D=\frac{80}{8}=10 \mathrm{~cm}$
So, the radius of the circle is $10\ cm.$

$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$
$\Rightarrow\ \angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$
Since $AOD$ is a straight line,
$\therefore\ \angle\text{COD}+\angle\text{AOC}=180^\circ$
$\therefore\ \angle\text{COD}+60^\circ=180^\circ$
$\therefore\ \angle\text{COD}=120^\circ$
$\Rightarrow\ \angle\text{CAO}=\frac{1}{2}\angle\text{COD}=\frac{1}{2}\times120^\circ=60^\circ$

Since angles in the same segment of a circle are equal.
$\angle\text{CDB}=\angle\text{BAC}$
That is , $\angle\text{ODB}=\angle\text{OAC}=50^\circ.$

$OA = OB = OC =$ Radius
So, $\angle\text{OAB}\angle\text{ABO}=40^\circ$
and, $\angle\text{OCB}=\angle\text{ABO}=30^\circ$
Thus, $\angle\text{ABC}=30+40=70^\circ$
So, $\angle\text{AOC}=70\times2=140^\circ$
{Angle subtended by arc at centre is twice of angle subtended at circumference}

Since $\triangle\text{BDC}$ is an equilateral traingle, $\angle\text{BAC}=60^\circ.$
Since $ABCD$ is a cyclic equilateral,
$\angle\text{BAC}+\angle\text{BDC}=180^\circ$
$\Rightarrow\ 60^\circ+\angle\text{BDC}=180^\circ$
$\Rightarrow\ \angle\text{BDC}=120^\circ$

We have:
$\angle\text{BOC}+\angle\text{BOA}+\angle\text{AOC}=360^\circ$
$\Rightarrow\angle\text{BOC}+100^\circ+90^\circ=360^\circ$
$\Rightarrow\angle\text{BOC}=(360^\circ-190^\circ)=170^\circ$
$\therefore\angle\text{BAC}=(\frac{1}{2}\times\angle\text{BOC})=(\frac{1}{2}\times170^\circ)=85^\circ$
$\Rightarrow\angle\text{BAC}=85^\circ$