Tag: Best NCERT Class 10th Maths Solutions

Q1

Question. Find the area of a sector of a circle with radius $6$ cm if angle of the sector is ${60}^{\circ }$.
Solution.

$$\text{Area}=\frac{\theta }{360}\pi r^2=\frac{60}{360}\pi (6{)}^2=\frac{1}{6}\pi \cdot 36=6\pi \mathrm{}$$

Using $\pi =\frac{22}{7}$:

$$\text{Area}=6\cdot \frac{22}{7}=\frac{132}{7}\approx \boxed{{\displaystyle 18.86{\text{cm}}^2}}\mathrm{}$$

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Q2

Question. Find the area of a quadrant of a circle whose circumference is $22$ cm.
Solution. Circumference $=2\pi r=22\Rightarrow r={\displaystyle \frac{11}{\pi }}$ With $\pi =\frac{22}{7}$ we get $r=3.5$.
Quadrant area $={\displaystyle \frac{1}{4}}\pi r^2={\displaystyle \frac{1}{4}}\pi (3.5{)}^2={\displaystyle \frac{1}{4}}\pi (12.25)=3.0625\pi$. With $\pi =\frac{22}{7}$,

$$\text{Area}\approx \boxed{{\displaystyle 9.62{\text{cm}}^2}}\mathrm{.}$$

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Q3

Question. The length of the minute hand of a clock is $14$ cm. Find the area swept by the minute hand in $5$ minutes.
Solution. In 60 minutes the hand sweeps ${360}^{\circ }$; in 5 minutes it sweeps ${30}^{\circ }$.
Area $={\displaystyle \frac{30}{360}}\pi (14{)}^2={\displaystyle \frac{1}{12}}\pi \cdot 196={\displaystyle \frac{49\pi }{3}}$. With $\pi =\frac{22}{7}$,

$$\text{Area}\approx \boxed{{\displaystyle 51.33{\text{cm}}^2}}\mathrm{.}$$

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Q4

Question. A chord of a circle of radius $10$ cm subtends a right angle at the centre. Find the area of the corresponding: (i) minor segment (ii) major sector. (Use $\pi =3.14$).
Solution. Here $\theta ={90}^{\circ }$, $r=10$
Sector area $={\displaystyle \frac{90}{360}}\pi r^2=\frac{1}{4}\pi (100)=25\pi =25\times 3.14=78.50{\text{cm}}^2\mathrm{}$
Triangle $\mathrm{△}OAB$ (central angle ${90}^{\circ }$) area $=\frac{1}{2}{r}^2\sin {90}^{\circ }=\frac{1}{2}\cdot 100\cdot 1=50{\text{cm}}^2\mathrm{}$

(i) Minor segment $=$ sector $-$ triangle $=78.50-50=\boxed{{\displaystyle 28.50{\text{cm}}^2}}\mathrm{}$
(ii) Major sector angle $=360-90={270}^{\circ }$. Major sector area $={\displaystyle \frac{270}{360}}\pi r^2=\frac{3}{4}\cdot 100\pi =75\pi =75\times 3.14=\boxed{{\displaystyle 235.50{\text{cm}}^2}}\mathrm{}$

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Q5

Question. In a circle of radius $21$ cm, an arc subtends an angle of ${60}^{\circ }$ at the centre. Find:
(i) length of the arc (ii) area of the sector formed by the arc (iii) area of the segment formed by the corresponding chord.
Solution. $r=21,\theta ={60}^{\circ },\pi =\frac{22}{7}\mathrm{}$

(i) Arc length $={\displaystyle \frac{\theta }{360}}\cdot 2\pi r={\displaystyle \frac{60}{360}}\cdot 2\pi \cdot 21={\displaystyle \frac{1}{6}}\cdot 42\pi =7\pi =7\frac{22}{7}=\boxed{{\displaystyle 22.00\text{ cm}}}\mathrm{}$

(ii) Sector area $={\displaystyle \frac{60}{360}}\pi r^2={\displaystyle \frac{1}{6}}\pi \cdot 441=73.5\pi$. With $\pi =\frac{22}{7}$ this is $\boxed{{\displaystyle 231.00{\text{cm}}^2}}\mathrm{}$

(iii) Triangle $OAB$ area $={\displaystyle \frac{1}{2}}{r}^2\sin {60}^{\circ }={\displaystyle \frac{1}{2}}\cdot 441\cdot \frac{\sqrt{3}}{2}={\displaystyle \frac{441\sqrt{3}}{4}}\approx 110.25\sqrt{3}$. Using $\sqrt{3}\approx 1.732$ gives triangle area $\approx 190.96{\text{cm}}^2$
Segment area $=$ sector $-$ triangle $\approx 231.00-190.96=\boxed{{\displaystyle 40.04{\text{cm}}^2}}\mathrm{}$

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Q6

Question. A chord of a circle of radius $15$ cm subtends an angle of ${60}^{\circ }$ at the centre. Find the areas of the corresponding minor and major segments.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$)
Solution. $r=15,\theta ={60}^{\circ }\mathrm{}$

Sector area $={\displaystyle \frac{1}{6}}\pi r^2={\displaystyle \frac{1}{6}}\cdot 3.14\cdot 225=117.75{\text{cm}}^2\mathrm{}$
Triangle area $=r^2\cdot \frac{\sqrt{3}}{4}=225\cdot \frac{1.73}{4}=97.31{\text{cm}}^2\mathrm{}$

Minor segment $=117.75-97.31=\boxed{{\displaystyle 20.44{\text{cm}}^2}}\mathrm{}$
Major segment $=$ total circle area $-$ minor segment

$=\pi r^2-$ minor segment$=3.14\cdot 225-20.44=706.50-20.44=\boxed{{\displaystyle 686.06{\text{cm}}^2}}\mathrm{}$

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Q7

Question. A chord of a circle of radius $12$ cm subtends an angle of ${120}^{\circ }$at the centre. Find the area of the corresponding segment.
(Use $\pi =3.14$ and $\sqrt{3}=1.73$).
Solution. $r=12,\theta ={120}^{\circ }\mathrm{}$

Sector area $={\displaystyle \frac{120}{360}}\pi r^2=\frac{1}{3}\cdot 3.14\cdot 144=150.72{\text{cm}}^2\mathrm{}$
Triangle area $=r^2\cdot \frac{\sqrt{3}}{4}=144\cdot \frac{1.73}{4}=62.28{\text{cm}}^2\mathrm{}$

Segment area $=150.72-62.28=\boxed{{\displaystyle 88.44{\text{cm}}^2}}\mathrm{}$

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Q8

Question. A horse is tied to a peg at one corner of a square grass field of side $15$ m by a rope of length $5$ m. (i) area of the part of field the horse can graze; (ii) increase in grazing area if rope were $10$ m long instead of $5$ m. (Use $\pi =3.14$).
Solution. The peg at a corner restricts the grazing inside the square to a quarter circle of radius equal to rope length (rope $<$ side).

(i) For rope $=5$ m: area $={\displaystyle \frac{1}{4}}\pi (5{)}^2={\displaystyle \frac{1}{4}}\cdot 3.14\cdot 25=19.625\approx \boxed{{\displaystyle 19.62{\text{m}}^2}}\mathrm{}$
(ii) For rope $=10$ m: area $={\displaystyle \frac{1}{4}}\pi (10{)}^2={\displaystyle \frac{1}{4}}\cdot 3.14\cdot 100=78.50{\text{m}}^2\mathrm{}$
Increase $=78.50-19.62=\boxed{{\displaystyle 58.88{\text{m}}^2}}\mathrm{}$

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Q9

Question. A brooch is made with silver wire in the form of a circle with diameter $35$ mm. The wire is also used in making $5$ diameters which divide the circle into $10$ equal sectors. Find: (i) total length of silver wire required. (ii) area of each sector of the brooch.
Solution. Diameter $d=35$, radius $r=17.5$ mm, $\pi =\frac{22}{7}$

(i) Total wire $=$ circumference $+$ $5$ diameters $=\pi d+5d=d(\pi +5)$. Numerically,

$$\text{circumference}=\pi d=\frac{22}{7}\cdot 35=110\text{ mm},5d=175\text{ mm}\mathrm{.}$$

So total $=110+175=\boxed{{\displaystyle 285.00\text{ mm}}}\mathrm{}$

(ii) Area of whole circle $=\pi r^2=\frac{22}{7}\cdot (17.5{)}^2=962.50\pi \mathrm{/}??$ (numeric) — but each sector is $1\mathrm{/}10$ of circle:

$$\text{area each}=\frac{1}{10}\pi r^2=\frac{1}{10}\cdot \frac{22}{7}\cdot (17.5{)}^2\approx \boxed{{\displaystyle 96.25{\text{mm}}^2}}\mathrm{.}$$

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Q10

Question. An umbrella has $8$ ribs equally spaced. Assuming umbrella is a flat circle of radius $45$ cm, find the area between two consecutive ribs.
Solution. Angle between consecutive ribs $={360}^{\circ }\mathrm{/}8={45}^{\circ }$

Area of sector

$$=\frac{45}{360}\pi (45{)}^2=\frac{1}{8}\pi \cdot 2025=\frac{2025\pi }{8}\mathrm{.}$$

With $\pi =\frac{22}{7}$,

$$\text{Area}=\frac{2025}{8}\cdot \frac{22}{7}\approx \boxed{{\displaystyle 795.54{\text{cm}}^2}}\mathrm{}$$

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Q11

Question. A car has two wipers which do not overlap. Each wiper has a blade of length $25$ cm sweeping through an angle of ${115}^{\circ }$. Find the total area cleaned at each sweep of the blades.
Solution. Area cleaned by one wiper $={\displaystyle \frac{115}{360}}\pi (25{)}^2$. Two wipers: double that.

$$\text{Total}=2\cdot \frac{115}{360}\pi \cdot 625.$$

With $\pi =\frac{22}{7}$ this evaluates to $\boxed{{\displaystyle 1254.96{\text{cm}}^2}}$

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Q12

Question. A lighthouse spreads red light over a sector of angle ${80}^{\circ }$ to a distance $16.5$ km. Find the area of sea warned. (Use $\pi =\frac{22}{7}$).
Solution. $r=16.5$ km, $\theta ={80}^{\circ }\mathrm{<span\; style="font-size:\; revert;\; font-family:\; var(--wp--preset--font-family--manrope);\; letter-spacing:\; -0.1px;">.</span>}$Area $={\displaystyle \frac{80}{360}}\pi (16.5{)}^2\approx \boxed{{\displaystyle 190.14{\text{km}}^2}}$

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Q13

Question. A round table cover has six equal designs as shown in figure. If radius $=28$, find the cost of making the designs at rate ₹0.35 per cm${}^2$. (Use $\sqrt{3}=1.7$).
Solution. Each design is the segment corresponding to $\theta ={60}^{\circ }$ (since $360\mathrm{/}6={60}^{\circ }$). For one segment:
Sector area $={\displaystyle \frac{60}{360}}\pi r^2={\displaystyle \frac{1}{6}}\cdot \frac{22}{7}\cdot {28}^2={\displaystyle \frac{1232}{3}}{\text{cm}}^2\approx 410.67$
Triangle $\mathrm{△}OAB$ (with central angle ${60}^{\circ }$) is equilateral of side $28$; area $={\displaystyle \frac{\sqrt{3}}{4}}\cdot {28}^2=196\sqrt{3}$. Using $\sqrt{3}=1.7$,

$$\text{triangle area}=196\times 1.7=333.20{\text{cm}}^2\mathrm{.}$$

Segment area $=$ sector $-$ triangle $\approx 410.67-333.20=77.47{\text{cm}}^2\mathrm{.}$ For six designs total area $=6\times 77.47=464.80{\text{cm}}^2\mathrm{.}$
Cost $=464.80\times 0.35=\boxed{{\displaystyle \text{₹}162.68}}\mathrm{}$

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Q14

Question. Tick the correct answer: Area of a sector of angle $p$ (degrees) of circle radius $R$ is — (options include the standard forms).
Solution. Formula is ${\displaystyle \frac{p}{360}}\pi R^2$. So the correct option is (C).

Q1

Question. From a point $Q$, the length of the tangent to a circle is $24$ cm and the distance of $Q$ from the centre is $25$ cm. The radius of the circle is
(A) $7$ cm (B) $12$ cm (C) $15$ cm (D) $24.5$ cm

Solution.
Let centre be $O$, point of tangency be $T$. In right triangle $OTQ$ (radius $OT$ ⟂ tangent at $T$):

$$O{Q}^2=O{T}^2+T{Q}^2\mathrm{}$$

So ${25}^2=r^2+{24}^2$. Hence

$$r^2=625-576=49\Rightarrow r=7\text{ cm}\mathrm{.}$$

Answer: (A) 7 cm.

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Q2

Question. In Fig. 10.11, if $TP$ and $TQ$ are two tangents to a circle with centre $O$ so that $\mathrm{\angle }POQ={110}^{\circ }$, then $\mathrm{\angle }PTQ$ is equal to (A) ${60}^{\circ }$ (B) ${70}^{\circ }$ (C) ${80}^{\circ }$ (D) ${90}^{\circ }$.

Solution.
The angle between the two tangents = ${180}^{\circ }$ (central angle between their points of contact). So

$$\mathrm{\angle }PTQ={180}^{\circ }-\mathrm{\angle }POQ={180}^{\circ }-{110}^{\circ }={70}^{\circ }\mathrm{.}$$

Answer: (B) ${70}^{\circ }$.

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Q3

Question. If tangents $PA$ and $PB$ from a point $P$ to a circle with centre $O$ are inclined to each other at angle ${80}^{\circ }$, then $\mathrm{\angle }POA$ is equal to (A) ${50}^{\circ }$ (B) ${60}^{\circ }$ (C) ${70}^{\circ }$ (D) ${80}^{\circ }$

Solution.
Let $\mathrm{\angle }APB={80}^{\circ }$. The radius $OP$ bisects the angle between the tangents, so $\mathrm{\angle }OPA={40}^{\circ }$. In right triangle $OAP$, angle at $A$ is ${90}^{\circ }$ (radius ⟂ tangent), so the three angles are ${90}^{\circ },{40}^{\circ },\mathrm{\angle }POA$. Thus

$$\mathrm{\angle }POA={180}^{\circ }-{90}^{\circ }-{40}^{\circ }={50}^{\circ }\mathrm{.}$$

Answer: (A) ${50}^{\circ }$.

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Q4

Question. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Solution.
Let diameter be $AB$ and tangents at $A$ and $B$ be $t_A$ and $t_B$. Radius $OA$ is collinear with $OB$ (they form the diameter). Tangent at $A$ is perpendicular to $OA$; tangent at $B$ is perpendicular to $OB$. Since $OA$ and $OB$ lie on the same straight line, the two perpendiculars are parallel. Hence tangents at the ends of a diameter are parallel.

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Q5

Question. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Solution.
Let circle centre $O$ and tangent at $P$ be line $t$. By Theorem 10.1, the radius $OP$ is perpendicular to the tangent at the point of contact $P$. Hence the perpendicular to the tangent at $P$ is exactly the radius $OP$, so it passes through the centre. ∎

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Q6

Question. The length of a tangent from a point $A$ at distance $5$ cm from the centre of the circle is $4$ cm. Find the radius of the circle.

Solution.
Let radius $=r$. In right triangle $OAT$:

$$O{A}^2=O{T}^2+A{T}^2\Rightarrow 5^2=r^2+4^2\mathrm{.}$$

So $r^2=25-16=9\Rightarrow r=3$

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Q7

Question. Two concentric circles are of radii $5$ cm and $3$ cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution.
Let the chord of larger circle touch the smaller at $P$. Distance from centre to the chord = radius of smaller circle $=3$. Half-chord length $=\sqrt{5^2-3^2}=\sqrt{25-9}=\sqrt{16}=4$. So full chord length $=2\times 4=8$ cm.

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Q8

Question. A quadrilateral $ABCD$ is drawn to circumscribe a circle (see Fig.10.12). Prove that

$$AB+CD=AD+BC\mathrm{.}$$

Solution.
Let the circle touch $AB,BC,CD,DA$ at $P,Q,R,S$ respectively. Tangent segments from a vertex to the circle are equal, so

$$AP=AS,BP=BQ,CQ=CR,DR=DS\mathrm{.}$$

Compute:

$$AB+CD=(AP+BP)+(CR+DR)=(AS+BQ)+(CQ+DS)\mathrm{.}$$

Using equalities $BP=BQ,CR=CQ,AP=AS,DR=DS$ we rearrange to get

$$AB+CD=(AS+DS)+(BQ+CQ)=AD+BC\mathrm{.}$$

Thus $AB+CD=AD+BC$

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Q9

Question. In Fig.10.13, $XY$ and $X^{\mathrm{\prime }}{Y}^{\mathrm{\prime }}$ are two parallel tangents to a circle with centre $O$ and another tangent $AB$ with point of contact $C$ intersects $XY$ at $A$ and $X^{\mathrm{\prime }}{Y}^{\mathrm{\prime }}$ at $B$. Prove that $\mathrm{\angle }AOB={90}^{\circ }$.

Solution. (coordinate / analytic proof — short and clean)
Place the circle with centre $O(0,0)$ and radius $r$. Take the two parallel tangents as the horizontal lines $y=r$ and $y=-r$. Then the tangent at a point $C(\cos \theta ,\sin \theta )$ has equation

$$x\cos \theta +y\sin \theta =r\mathrm{.}$$

Its intersection $A$ with $y=r$ is at

$$x_A=\frac{r-r\sin \theta }{\cos \theta }=\frac{r(1-\sin \theta )}{\cos \theta },A=(\frac{r(1-\sin \theta )}{\cos \theta },r)\mathrm{.}$$

Its intersection $B$ with $y=-r$ is

$$x_B=\frac{r+r\sin \theta }{\cos \theta }=\frac{r(1+\sin \theta )}{\cos \theta },B=(\frac{r(1+\sin \theta )}{\cos \theta },-r)\mathrm{.}$$

Compute dot product of vectors $\overrightarrow{OA}$and $\overrightarrow{OB}$:

$$\overrightarrow{OA}\cdot \overrightarrow{OB}=\frac{r^2(1-{\sin }^2\theta )}{{\cos }^2\theta }+(r)(-r)=\frac{r^2{\cos }^2\theta }{{\cos }^2\theta }-r^2=r^2-r^2=0.$$

So $OA\perp OB$, therefore $\mathrm{\angle }AOB={90}^{\circ }$.

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Q10

Question. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Solution.
Let the tangents from external point $P$ touch the circle at $A$ and $B$. Then (standard result)

$$\mathrm{\angle }APB={180}^{\circ }-\mathrm{\angle }AOB,$$

because each tangent is perpendicular to the radius at the point of contact; triangle $OAP$ and $OBP$ give $\mathrm{\angle }OPA=\frac{1}{2}\mathrm{\angle }APB$ and $\mathrm{\angle }OPA+\mathrm{\angle }POA+{90}^{\circ }={180}^{\circ }$ etc. So $\mathrm{\angle }APB+\mathrm{\angle }AOB={180}^{\circ }$. Hence they are supplementary. ∎

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Q11

Question. Prove that the parallelogram circumscribing a circle is a rhombus.

Solution.
If a parallelogram circumscribes a circle, then by Q8 we have for the circumscribed quadrilateral $ABCD$:

$$AB+CD=AD+BC\mathrm{.}$$

But in a parallelogram $AB=CD$ and $BC=AD$. Substitute to get:

$$AB+AB=AD+AD\Rightarrow 2AB=2AD\Rightarrow AB=AD\mathrm{.}$$

Thus two adjacent sides are equal; in a parallelogram equal adjacent sides imply all four sides are equal. Therefore the parallelogram is a rhombus. ∎

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Q12

Question. A triangle $ABC$ is drawn to circumscribe a circle of radius $4$ cm such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8$ cm and $6$ cm respectively (see Fig.10.14). Find the sides $AB$ and $AC$.

Solution.
Let the circle touch $AB,BC,CA$ at $F,D,E$ respectively. Tangent segments from same vertex are equal: let $AF=AE=x$. From given,

$$BF=BD=8,CE=CD=6.$$

Thus

$$AB=AF+BF=x+8,AC=AE+CE=x+6,BC=8+6=14.$$

Let semiperimeter $s={\displaystyle \frac{AB+BC+CA}{2}}={\displaystyle \frac{(x+8)+14+(x+6)}{2}}=x+14$. Area of triangle equals $r\cdot s$ where $r=4$, so $\mathrm{\Delta }=4(x+14)$. By Heron,

$$\mathrm{\Delta }=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{(x+14)\cdot (x)\cdot 8\cdot 6}=\sqrt{48x(x+14)}\mathrm{.}$$

Square both expressions:

$$[4(x+14){]}^2=48x(x+14)\Rightarrow 16(x+14{)}^2=48x(x+14)\mathrm{.}$$

Divide by $(x+14)$:

$$16(x+14)=48x\Rightarrow 16x+224=48x\Rightarrow 224=32x\Rightarrow x=7.$$

So

$$AB=x+8=7+8=15\text{ cm},AC=x+6=7+6=13\text{ cm}\mathrm{.}$$

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Q13

Question. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution.
Let $ABCD$ be a cyclic quadrilateral circumscribing a circle (i.e., circle is inscribed in the quadrilateral), and let the circle center be $O$. Let the circle touch sides $AB,BC,CD,DA$ at $P,Q,R,S$ respectively. Because tangents from the same vertex are equal:

$$AP=AS,BP=BQ,CQ=CR,DR=DS\mathrm{}$$

Consider central angles subtended by chords $AB$ and $CD$, i.e. $\mathrm{\angle }AOB$ and $\mathrm{\angle }COD$. The arcs corresponding to chords $AB$ and $CD$ together make the whole circle because the contact points split the circle into four arcs whose pairwise sums correspond to opposite sides. More directly, one can show (using equality of tangent segments) that the length of arc $AB$ + length of arc $CD$ = full circumference, hence the sum of the central angles is ${360}^{\circ }$. But each central angle measures twice the angle subtended by the chord at the centre, so for chords opposite each other the central angles add to ${360}^{\circ }$; since we consider the interior angles subtended at the centre by the segments, we get

$$\mathrm{\angle }AOB+\mathrm{\angle }COD={360}^{\circ }-(\text{reflex angles})\Rightarrow \mathrm{\angle }AOB+\mathrm{\angle }COD={180}^{\circ }\mathrm{.}$$

(Equivalently, one may give a more metric proof: show arcs $AB$ and $CD$ are supplementary because tangency-length equalities force the arcs cut off by the contact points to pair suitably; hence central angles over those arcs add to ${180}^{\circ }$)

Thus opposite sides subtend supplementary central angles.