SELINA Solution Class 9 Chapter 17 Circle Exercise 17C

Question 1

In the given figure, an equilateral triangle ABC is inscribed in a circle with center O.
Find: (i) ∠BOC
(ii) ∠OBC

Sol:

In the given figure, ABC is an equilateral triangle.

Hence all the three angles of the triangle will be equal to 60°
i.e. ∠A = ∠B = ∠C = 60°

As the triangle is an equilateral triangle, BO and CO will be the angle bisectors of B and C respectively.

Hence ∠OBC = ${\displaystyle \frac{\text{∠ABC}}{2}}$ 
                     = 30°
and as given in the figure we can see that OB and OC are the radii of the given circle.
Hence they are of equal length.

The ΔOBC is an isosceles triangle with OB = OC
In ΔOBC,
∠OBC = ∠OCB as they are angles opposite to the two equal sides of an isosceles triangle.
Hence, ∠OBC = 30° and ∠OCB = 30°
Since the sum of all angles of  a triangle is 180°

Hence in triangle OBC, ∠OCB + ∠OBC + ∠BOC + BOC = 180°
30° + 30° + ∠BOC= 180°
60° + BOC = 180°
∠BOC = 180° - 60°
∠BOC = 120°
Hence ∠BOC =120° and ∠OBC =30°

Question 2

In the given figure, a square is inscribed in a circle with center O.
Find:
(i) ∠BOC
(ii) ∠OCB
(iii) ∠COD
(iv) ∠BOD
Is BD a diameter of the circle?

Sol:

In the given figure we can extend the straight line OB to BD and CO to CA

Then we get the diagonals of the square which intersect each other at 90 by the property of Square.

From the above statement, we can see that
∠COD = 90°

The sum of the angle ∠BOC and ∠COD is 180° as BD is a straight line.
Hence ∠BOC + ∠OCD = ∠BOD = 180°
∠BOC + 90° = 180°
∠BOC + 180° - 90°
∠BOC = 90°

We can see that the OCB is an isosceles triangle with sides OB and OC of Equal length as they are the radii of the same are.

In ΔOCB,
∠OBC = ∠OCB as they are opposite angles to the two equal sides of an isosceles triangle.

Sum of all the angles of a triangle is 180°
so, ∠OBC + ∠OCB + ∠BOC =180°
∠OBC + ∠OBC + 90° = 180° as, ∠OBC = ∠OCB
2∠OBC = 180° - 90°
2∠OBC = 90°
2∠OBC = 45°
as ∠OBC = ∠OCB So,
∠OBC = OCB = 45°

Yes BD is the diameter of the order.

Question 3

In the given figure, AB is a side of regular pentagon and BC is a side of regular hexagon.
(i) ∠AOB
(ii) ∠BOC
(iii) ∠AOC
(iv) ∠OBA
(v) ∠OBC
(vi) ∠ABC

Sol:

As given that AB is the side of a pentagon the angle subtended by each arm of the pentagon at the center of the circle is = ${\displaystyle \frac{360°}{5}}$ = 72°

Thus angle ∠AOB = 72°

Similarly, as BC is the side of a hexagon hence the angle subtended by BC at the center is = ${\displaystyle \frac{360°}{6}}$ i.e. 60°
∠BOC = 60°

Now ∠AOC = ∠AOB + ∠BOC =72° + 60° = 132°

The triangle thus formed, ΔAOB is an isosceles triangle with OA = OB as they are radii of the same circle.

Thus ∠OBA = ∠BAO as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°
so, ∠AOB + ∠OBA + ∠BAO = 180°
⇒ 2∠OBA + 72° = 180° as, ∠OBA = ∠BAO
⇒ 2∠OBA = 180° - 72°
⇒ 2∠OBA = 180°
⇒ 2∠OBA =54° 
as, ∠OBA = ∠BAO So,
∠OBA = ∠BAO = 54°

The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same are.

Thus ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°
so, ∠BOC + ∠OBC + ∠OCB = 180°
2∠OBC + 60° = 180° as , ∠OBC = ∠OCB
2∠OBC = 180° - 60°
2∠OBC = 120°
∠OBC = 60°
as ∠OBC = ∠OCB
So, ∠OBC = ∠OCB = 60°
∠ABC = ∠OBA + ∠OBC = 54° + 60°= 114°

Question 4

In the given figure, arc AB and arc BC are equal in length. If ∠AOB = 48°, find:
(i) ∠BOC
(ii) ∠OBC
(iii) ∠AOC
(iv) ∠OAC

Sol:

We know that the arc of equal lengths subtends equal angles at the center.

hence ∠AOB = ∠BOC = 48°
Then ∠AOC = ∠AOB + ∠BOC = 48° + 48° = 96°
The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same circle.

Thus ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°.
So, ∠BOC + ∠OBC + ∠OCB = 180°   
 2∠OBC + 48° = 180°              as ∠OBC = ∠OCB
 2∠OBC = 180° - 48°
 2∠OBC = 132°
∠OBC = 66°
as ∠OBC = ∠OCB 
So, ∠OBC = ∠OCB = 66°

The triangle thus formed, ΔAOC is an isosceles triangle with OA = OC as they are radii of the same circle.
Thus ∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°.
So, ∠COA + ∠OAC + ∠OCA = 180°
2∠OAC + 96° = 180°                 as, ∠OAC = ∠OCA 
2∠OAC = 180° - 96°
2∠OAC = 84° 
∠OAC = 42°
as ∠OCA = ∠OAC
So, ∠OCA = ∠OAC = 42°.

Question 5

In the given figure, the lengths of arcs AB and BC are in the ratio 3:2. If ∠AOB = 96°,
find: (i) ∠BOC (ii) ∠ABC

Sol:

We know that for two arcs are in ratio 3: 2 then
∠AOB: ∠BOC = 3: 2
As give ∠AOC = 96°
So, 3x = 96
       x = 32
There ∠BOC = 2 x 32 = 64°
The triangle thus formed, ΔAOB is an isosceles triangle with OA = OB as they are radii of the same circle.

Thus ∠OBA = ∠BAO as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOB + ∠OBA + ∠BAO = 180°
2∠OBA + 96° = 180°       as,  ∠OBA = ∠BAO
2∠OBA = 180° - 96° 
2∠OBA = 84°
∠OBA = 42°
as, ∠OBA = ∠BAO So,
∠OBA = ∠BAO = 42°

The triangle thus formed, ΔBOC is an isosceles triangle with OB = OC as they are radii of the same circle.
Thus ∠OBC = ∠OCB as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠BOC + ∠OBC + ∠OCB = 180°
2∠OBC + 64° = 180°          as, ∠OBC = ∠OCB 
2∠OBC = 180° - 64° 
2∠OBC =  116° 
∠OBC = 58°
As ∠OBC = ∠OCB   So,
∠OBC = ∠OCB = 58° 
∠ABC = ∠BOA + ∠OBC = 42°+ 58° = 100°

Question 6

In the given figure, AB = BC = DC and ∠AOB = 50°.
(i) ∠AOC
(ii) ∠AOD
(iii) ∠BOD
(iv) ∠OAC
(v) ∠ODA

Sol:

Since arc AB and BC are equal.
So, ∠AOB = ∠BOC = 50°
Now,
∠AOC = ∠AOB + ∠BOC = 50° + 50° = 100°
As arc AB, arc BC and arc CD so, 
∠AOB = ∠BOC = ∠COD = 50°
∠AOD = ∠AOB + ∠BOC + ∠COD = 50° + 50° + 50° = 150°
Now, ∠BOD = ∠BOC + ∠COD 
∠BOD = 50° + 50° 
∠BOD = 100°

The triangle thus formed, ΔAOC is an isosceles triangle with OA = OC as they are radii of the same circle.
 Thus ∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOC + ∠OAC + ∠OCA = 180°
2∠OAC + 100° = 180°     as, ∠OAC = ∠OCA
2∠OAC = 180° - 100° 
2∠OAC = 80°
∠OAC = 40°

as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 40°

The triangle thus formed, ΔAOD is an isosceles triangle with OA = OD as they are radii of the same circle.
Thus, ∠OAD = ∠ODA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOD + ∠OAD + ∠ODA = 180°
2∠OAD + 150° = 180°         as, ∠OAD = ∠ODA
2∠OAD = 180° - 150° 
∠OAD = 30°

as ∠OAD = ∠ODA    So,
∠OAD = ∠ODA  = 15°.

Question 7

In the given figure, AB is a side of a regular hexagon and AC is a side of a regular eight-sided polygon.
Find:
(i) ∠AOB
(ii) ∠AOC
(iii) ∠BOC 
(iv) ∠OBC

Sol:

As AB is the side of a hexagon so the
∠AOB = ${\displaystyle \frac{360°}{6}}$ = 60°

AC is the side of an eight-sided polygon so,
∠AOC = ${\displaystyle \frac{360°}{8}}$ = 45°

From the given figure we can see that:
∠BOC = ∠AOB + ∠AOC 
⇒ 60° + 45° = 105°
Again, from the figure, we can see that ∠BOC is an isosceles triangle with sides BO = OC as they are the radii of the same circle.
Angles ∠OBC = ∠OCB    as they are opposite angles to the equal sides of an isosceles triangle.

Sum of all the angles of a triangle is 180°
∠OBC + ∠OCB + ∠BOC = 180°
2∠OBC + 105° = 180°                    as, ∠OBC = ∠BOC
2∠OBC = 180° - 105°
2∠OBC = 75°
∠OBC = 37.5° = 37°30'
As, ∠OBC = ∠BOC
∠OBC = ∠BOC = 37.5° = 37°30'.

Question 8

In the given figure, O is the center of the circle and the length of arc AB is twice the length of arc BC. If ∠AOB = 100°,
find: (i) ∠BOC (ii) ∠OAC

Sol:

We know that when two arcs are in ratio 2: 1 then the subtended by them is also in ratio 2: 1
As given arc AB is twice the length of arc BC.
Therefore, arc AB: arc BC = 2: 1
Hence, ∠AOB: ∠BOC = 2: 1

Now given that ∠AOB = 100°.
So, ∠BOC = ${\displaystyle \frac{1}{2}\angle \text{AOB}=\frac{1}{2}\times 100°=50°}$

Now, ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.
The triangle thus formed, ∠AOC is an isosceles triangle with OA = OC as they are radii of the same circle.
Thus,
∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°.
So, ∠COA + ∠OAC + ∠OCA = 180°
2∠OAC + 150° = 180° as, ∠OAC = ∠OCA 
2∠OAC = 180° - 150°
2∠OAC = 30°
∠OAC = 15°
as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 15°.