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Science > Mathematics > Trigonometry > You are Here |

#### Example – 01:

- Find the length of the arc of circle of diameter 10 cm, if the arc is subtending an angle of 36° at the centre.
**Given:**Diameter = 10 cm, radius = r = 10/2 = 5 cm, angle subtended = 36° = 36 x π/180 = (π//5)^{c}**To Find:**Length of arc = S = ?**Solution:**

length of arc is given by

S = r θ = 5 x π//5 = π cm

**Ans:** The length of arc is π cm

#### Example – 02:

- Find the length of the arc of circle which subtends an angle of 108° at the centre, if theradius of circle is 15 cm.
**Given:**radius = r = 15 cm, angle subtended = 108° = 108 x π/180 = (3π//5)^{c}**To Find:**Length of arc = S = ?**Solution:**

length of arc is given by

S = r θ = 15 x 3π//5 = 9π cm

**Ans:** The length of arc is 9π cm

#### Example – 03:

- The radius of circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length equal to the length of radius.
**Given:**radius = r = 9 cm, length of chord = r**To Find:**Length of arc = S = ?**Solution:**

Thus ΔOAB is equilateral triangle

angle subtended = 60° = 60 x π/180 = (π//3)^{c}

length of arc is given by

S = r θ = 9 x π//3 = 3π cm

**Ans:** The length of arc is 3π cm

#### Example – 04:

- In a circle of diameter 40 cm, the length of chord is 20 cm. Find the length of minor arch of the chord.

**Given:**diameter = 4o cm, radius = r = 20 cm, length of chord = 20 cm**To Find:**Length of arc = S = ?**Solution:**

Thus ΔOAB is equilateral triangle

angle subtended = 60° = 60 x π/180 = (π//3)^{c}

length of arc is given by

S = r θ = 20 x π//3 = 20π/3 cm

**Ans:** The length of minor arc is 20π/3 cm

#### Example – 05:

- A pendulum of 14 cm long oscillates through an angle of 18°. Find the length of the path described by its extremity..
**Given:**radius = r = 14 cm, angle subtended = 18° = 18 x π/180 = (π//10)^{c}**To Find:**Length of arc = S = ?**Solution:**

Length of path i.e. length of arc is given by

S = r θ = 14 x π/10 = 1.4π cm

**Ans:** The length of arc is 1.4π cm

#### Example – 06:

- Find in radians and degrees the angle subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of circle is 25 cm.
**Given:**radius = r = 25 cm, Length of arc = 15 cm**To Find:**angle subtended at the centre = θ = ?**Solution:**

length of arc is given by

S = r θ

∴ θ = S/r = 15/25 = (3/5)^{c}

(3/5)^{c }= (3/5) x (180/π) = (108/π)°

**Ans:** The angle subtended at the centre is (3/5)^{c }or (108/π)°

#### Example – 07:

- Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm. Take ( π = 22/7)
**Given:**Length of arc = S = 37.4 cm, angle subtended = 60° = 60 x π/180 = (π//3)^{c}**To Find:**Length of arc = S = ?**Solution:**

Length of path i.e. length of arc is given by

S = r θ

∴ r =S/θ = 37.4/(π//3) = (37.4 x 3) / (22/7)

∴ r = (37.4 x 3 x 7) / 22 = 35.7

**Ans:** The length of arc is 35.7 cm

#### Example – 08:

- A wire of length 10 cm is bent so as to form an arc of a circle of radius 4 cm. What is the angle subtended at the centre in degrees.
**Given:**radius = r = 4 cm, length of wire = length of arc = 10 cm**To Find:**angle subtended at the centre = θ = ?**Solution:**

length of arc is given by

S = r θ

∴ θ = S/r = 10/4 = (2.5)^{c}

(2.5)^{c }= (2.5) x (180/π) = (450/π)°

**Ans:** The angle subtended at the centre is (2.5)^{c }or (450/π)°

#### Example – 09:

- Δ PQR is an equilateral triangle with side 18 cm. A circle is drawn on segment QR as diameter. Find the length of the arc of this circle intercepted within the triangle.
**Given:**Side of equilateral triangle = 18 cm, radius = 18/2 = 9 cm**To Find:**length of the arc of circle intercepted = S = ?**Solution:**

As Δ PQR is equilateral triangle, its each angle is 60°

Hence the triangles Δ QOE and Δ ROD are also equilateral triangles

∠ EOQ = ∠ ROD = 60°

EOD = 60° = θ

length of arc is given by

S = r θ = 9 x π//3 = 3π cm

**Ans:** The length of arc is 3π cm

#### Example – 10:

- Two arcs of the same length subtends angle 60° and 75° at the centres of the circles. What is the ratio of the radii of the two circles.
**Given:**Angles subtended, θ_{1}= 60° and θ_{2}= 75°**To Find:**Ratio of radii = r_{1}/r_{2}= ?**Solution:**

length of arc is given by S = r θ

For the first arc S_{1} = r_{1}θ_{1} ………. (1)

For the second arc S_{2} = r_{2}θ_{2} ………. (2)

Now length of two arcs is the same

S_{1} = S_{2}

∴ r_{1}θ_{1} = r_{2}θ_{2}

∴ r_{1}/r_{2} = θ_{2}/ θ_{1}

∴ r_{1}/r_{2} = 75°/ 60° = 5/4

**Ans:** The ratio of radii is 5:4

#### Example – 11:

- Two arcs of the same length subtends angle 65° and 110° at the centres of the circles. What is the ratio of the radii of the two circles.
**Given:**Angles subtended, θ_{1}= 65° and θ_{2}= 110°**To Find:**Ratio of radii = r_{1}/r_{2}= ?**Solution:**

length of arc is given by S = r θ

For the first arc S_{1} = r_{1}θ_{1} ………. (1)

For the second arc S_{2} = r_{2}θ_{2} ………. (2)

Now length of two arcs is the same

S_{1} = S_{2}

∴ r_{1}θ_{1} = r_{2}θ_{2}

∴ r_{1}/r_{2} = θ_{2}/ θ_{1}

∴ r_{1}/r_{2} = 110°/ 65° = 22/13

**Ans:** The ratio of radii is 22:13

#### Example – 12:

- A train is running on a circular track of radius 1 km at the rate of 36 km per hour. Find the angle to the nearest minute, through which it will turn in 30 seconds.
**Given:**Radius of the arc = r = 1 km = 1000 m, Speed of train = v = 36 km per hour = 36 x 1000/3600 = 10 m/s, time taken = t = 30 s.**To Find:**Angle through which the train turns = θ = ?**Solution:**

Distance covered by train i.e. length of arc = speed x time = 10 x 30 = 300 m

S = r θ

∴ θ = S/r = 300/1000 = (0.3)^{c}

∴ θ = (0.3)^{c }= (0.3) x (180/π) = (54/3.142)° = 17.19°

∴ θ = 17° + 0.19° = 17° + 0.19 x 60′ = 17° + 11′ = 17°,11′

**Ans:** The train will turn through 17°,11′

#### Example – 13:

- A train is running on a circular track of radius 1500 m at the rate of 66 km per hour. Find the angle to the in radian, through which it will turn in 10 seconds.
**Given:**Radius of the arc = r = 1500 m, Speed of train = v = 66 km per hour = 66 x 1000/3600 = 55/3 m/s, time taken = t = 10 s.**To Find:**Angle through which the train turns = θ = ?**Solution:**

Distance covered by train i.e. length of arc = speed x time = (55/3) x 10 = 550/3 m

S = r θ

∴ θ = S/r = (550/3)/1500 = 550/4500 = (11/90)^{c}

**Ans:** The train will turn through (11/90)^{c}

#### Example – 14:

- A horse is tied to a post by a rope. If the horse moves along a circular path, always keeping the rope tight and describes 88 m when it traces an angle of 72° at the centre, find the length of the rope. Take π = 22/7.
- Given: central angle = θ = 72° = 72 x π/180 = (2π/5)
^{c}, arc length = S = 88 m - To Find: Length of rope = r = ?

S = r θ

∴ r = S/θ = 88/(2π/5) = 220/π = 220 x 7/22 = 70 m

**Ans:** The length of rope is 70 m.

#### Example – 15:

- If the perimeter of a sector of a circle is four times the radius of the circle, find the central angle of corresponding sector in radians.
**Given:**Perimeter = 4 x radius = 4r**To find:**Central angle = θ = ?

Perimeter of sector = r + r + s = 4r

∴ 2r + r θ = 4r

∴ r θ = 2r

∴ θ = 2^{c}

**Ans:** The central angle is 2^{c}

Science > Mathematics > Trigonometry > You are Here |

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