Maharashtra Board Class 11 Maths Solution Chapter 1 – Angle and its Measurement
Balbharati Maharashtra Board Class 11 Maths Solution Chapter 1: Angle and its Measurement. Marathi or Maths Medium Students of Class 11 get here Angle and its Measurement full Exercise Solution.
|
Std |
Maharashtra Class 11 |
| Subject |
Mathematics & Statistics (Part 1) |
|
Chapter |
1 |
| Chapter Name |
Angle and its Measurement |
Part 1 Chapter 1 – Angle and its Measurement
Exercise – 1.1
(Q1) (A) Determine which of the following pairs of angles are co-terminal.
(i) 210°, −150°
(ii) 360°, −30°
(iii) −180°, 540°
(iv) −405°, 675°
(v) 860°, 580°
(vi) 900°, −900°
Solution:
(i) 210°, -150°
210° – (-150°) = 210° + 150° = 360° Multiple of 360°
∴ The angles are co-terminal
(ii) 360°, −30°
Solution:
360°, −30°
360° – (-30°) = 360° + 30° = 390° ≠ multiply of 360°
∴ The angles are not co-terminal
(iii) −180°, 540°
540 ° – (-180°) = 540° + 180° = 720° = 2×360° = multiple of 360°
∴ The angles are co-terminal
(iv) −405°, 675°
675° – (-405°) = 675° + 405° = 1080° = 3×360° = multiple of 360°
∴ The angles are co-terminal
(v) 860°, 580°
860° – 580° = 280° ≠ multiple of 360°
∴ The angle are not co-terminal
(vi) 900°, −900°
900° – (-900°) = 900° + 900° = 1800° = 5×360° = multiple of 360°
∴ The angles are co-terminal
(B) Draw the angles of the following measures and determine their quadrants.
(i) –140°
(ii) 250°
(iii) 420°
(iv) 750°
(v) 945°
(vi) 1120°
(vii) –80°
(viii) –330°
(ix) –500°
(x) –820°
Solution:
(Q2) Convert the following angles in to radian.
(i) 85°
(ii) 250°
(iii) −132°
(iv) 65°30′
(v) 75°30′
(vi) 40°48′
Solution:
(i) 85°
85° × π/180° = (17/36 π)C
(ii) 250°
250° × π/180° = (25π/18)C
(iii) -132°
– (132° × π/180°) = (-11/15 π)C
(iv) 65°30′
65°30′ = 65° + (30/60)° = 65.5°
65.5° × π/180° = (131/360 π)C
(v) 75°30′
75°30′ = 75° + (30/60)°
= 75.5°
75.6 × π/180° = (151/360 π)C
(vi) 40°48′
40°48′ = 40° + (48/60)°
= 40°.8°
40.8° × π/180° = (17π/75)C
(Q3) Convert the following angles in degree.
(i) 7πC/12
(ii) -5πC/3
(iii) 5C
(iv) 11πC/18
(v) (-1/4)C
Solution:
(i) 7πC/12
7π/12 × 180°/π = 180°
(ii) -5πC/3
– (5π/3 × 180°/π) = -300°
(iii) 5C
5 × 180°/π = 286.624204°
= 286° + (0.624204 × 60)′
= 286° 37′ + (0.45224 × 60)″
= 286° 37′ 27.1344″
= 286° 37′ 27″
(iv) 11πC/18
11π/18 × 180°/π = 110°
(v) (-1/4)C
= – (1/4 × 180°/ π = -45°/3.14
= -14.3312102°
= – {14° + (0.3312102 × 60)′}
= – {14° 19.872612′}
= – {14° 19′ + (0.872612 × 60)″}
= – {14° 19′ – 52.35672″}
= -14°19′52″
(Q4) Express the following angles in degree, minute and second.
(i) (183.7)°
(ii) (245.33)0
(iii) (1/5)C
Solution:
(i) (183.7)°
183.7° = 183° + (0.7×60)′
= 183° 42′
(ii) (245.33)0
245.33° = 245° + (0.33×60)′
= 245° 19.8′
= 245° 19′ + (0.8×60)″
= 245o19’48”
(iii) (1/5)C
1/5 × 180/π = 11.4649682°
= 11° + (0.4649682×60)′
= 11° 27.868092′
= 11° 27′ + (0.898092 × 60)″
= 11° 27′53.88552″
= 11° 27′ 54″
(Q5) In △ABC, if m ∠A = 7πC/36,
m ∠B = 120°, find m ∠C in degree and radian.
Solution:
m ∠A = 7πC/36 = 7π/36 × 180°/π = 35°
m ∠B = 120° = 120° × π/180° = 2/3 π
∴ m ∠C + m ∠A + m ∠B = 180°
=> m ∠C = 180° – 120° – 35° = 25°
∴ m ∠C = 25° × π/180° = 5/36 πC
(Q6) Two angles of a triangle are 5πC/9 and 5πC. Find the degree and radian measure of third angle.
Solution:
Given angles, 5πC/9 = 5π/9 × 180°/π = 100°
5πC/18 = 5π/18 × 180°/π = 50°
∴ Third angle = 180° – 100° – 50°
= 30°
= 30° × π/180°
= πC/60°
(Q7) In a right angled triangle, the acute angles are in the ratio 4:5. Find the angles of the triangle in degree and radian.
Solution:
Let, the two angles be 4k & 5k.
∵ It’s a right angled triangle,
∴ 4k + 5k = 90°
= k = 10°
∴ The angles are 40° and 50°
∴ 40° = 40° × π/180° = 2πC/9 and 50° = 50° × π/180° = 5πC/18
(Q8) The sum of two angles is 5πC and their difference is 60°. Find their measures in degree.
Solution:
5πC = 5π × 180°/π = 900°
Let, the two angles be x & y
A/Q, x + y = 900°
x – y = 60°
∴ 2x = 960° = x = 480°
2y = 840 = y = 420°
(Q9) The measures of the angles of a triangle are in the ratio 3:7:8. Find their measures in degree and radian.
Solution:
Let, the angles be 3K, 7K and 8K
∴ 3K + 7K + 8K = 180°
= 18K = 180°
= K = 10°
∴ The angles are: 30° = 30° × π/180° πC/6
70° = 70° × π/180° = 7πC/18
80° = 80° × π/180° = 4πC/9
(Q10) The measures of the angles of a triangle are in A.P. and the greatest is 5 times the smallest (least). Find the angles in degree and radian.
Solution:
Let, the angles of the triangle be a – d, a, a + d
A/Q, a + d = 5 (a – d)
=> 4a = 6d
=> a = 3/2 d —— (i)
Also, a – d + a + a + d = 180°
=> 3a = 180°
=> a = 60°
∴ (i) => d = 2/3 × 60° = 40°
∴ The angles are – 20°, 60°, 100°
20° = 20° × π/180° = πC/9
60° = 60° × π/180° = πC/3
100° = 100° × π/180° = 5πC/9
(Q11) In a cyclic quadrilateral two adjacent angles are 40° and πC/3. Find the angles of the quadrilateral in degree.
Solution:
Given angles,
40° and πC/3 = π/3 × 180/π
= 60o
We know, opposite angles of cyclic quadrilateral are supplementary.
Let, The angles be ∠A, ∠B, ∠C and ∠D.
∴ ∠A = 40°, ∠B = 60°
∴ ∠A + ∠C = 180°
=) ∠C = 140°
and ∠B + ∠D = 180°
=) ∠D = 120°
(Q12) One angle of a quadrilateral has measure 2πC and the measures of other three angles are in the ratio 2:3:4. Find their measures in degree and radian.
Solution:
Given angle, 2πC/5 = 2π/5 × 180°/π
= 72°
Let, the angles be 2K, 3K & 4K
∴ 2K + 3K + 4K + 72° = 360°
=) 9K = 288°
=) K = 32°
∴ The angles are: 64° = 64° × π/180° = 16/45 πC
& 96° = 96° × π/180° = 8/15 πC
& 128° = 128° × π/180° = 32/45 πC
(Q13) Find the degree and radian measure of exterior and interior angle of a regular
(i) Pentagon
(ii) Hexagon
(iii) Septagon
(iv) Octagon
Solution:
(i) Pentagon:
A Pentagon has 5 sides and exterior angle each exterior angle = 360°/5 = 72° = 72° × π/180° = 2/3 πC
Interior angle + exterior angle = 180°
= Each interior angle = 180° – 72°
= 108°
= (108° × π/180°)C
= 3/5 πC
(ii) Hexagon
A Hexagon has 6 side and 6 exterior angles each exterior angle = 360°/6 = 60° = 60° × π/180°
= πC/3
Each interior angle = 180° – 60°
= 120° = 120° × π/180°
= 2πC/3
(iii) Septagon
A Septagon has 7 sides and 7 exterior angles each exterior angle = 360°/7 = 51.4° (approve)
= 360°/7 × π/180° = 2πC/7
Each interior angle = 180° – 51.4
= 128.6°
= 900°/7 × π/180°
= 5πC/7
(iv) Octagon
A Octagon has 8 sides and 8 exterior angles each exterior angle = 360°/8 = 45°
= 45° × π/180°
= πC/4
Each interior angle = 180° – 45°
= 135°
= 135° × π/180°
= 3/4 πC
(Q14) Find the angle between hour-hand and minute-hand in a clock at
(i) ten past eleven
(ii) twenty past seven
(iii) thirty five past one
(iv) Quarter to six
(v) 2:20
(vi) 10:10
Solution:
(i) Ten past eleven
We know, angle between two clock mark is 30°
∴ Angle between hour and minute hand is less than (3×30°) = 90°
In one minute, hour hand moves by 1/2 o
∴ In 10 minutes, hour hand moves by 10°/2 = 5°
∴ Angle between hour & minute hand = 90° – 5° = 85°
(ii) Twenty past seven
We know, angle between two clock mark is 30°
∴ Angle between hour & minute hand is more than (3×30°) = 30°
In one minute, hour hand moves by 1°/2
∴ In 20 minutes, hour hand moves by 20°/2 = 10°
∴ Angle between hour & minute hand = 90° + 10° = 100°
(iii) Thirty five past one
We know, angle between two clock mark is 30°
∴ Angle between hour & minute hand is less than (6×30°) = 180°
In one minute, hour hand moves by 1/2 o
In 35 minutes, hour hand moves by 35°/2 = 17.5°
∴ Angle between hour & minute hand = 180° – 17.5° = 162.5° = 162° 30’
(iv) Quarter to six
We know, angle between two clock mark is 30°
∴ Angle between hour & minute hand is more than (4 × 30°) = 120°
In one minute, hour hand moves by 1/2 o
In 45 minutes, hour hand moves by 45°/2 = 22.5°
∴ Angle between hour & minute hand = 120° – 22.5° = 97.5° = 97°30’
(v) 2:20
We know, angle between two clock mark is 30°
∴ Angle between hour & minute hand is less than (2×30°) = 60°
In one minute, hour hand moves by 1/2 o
In 20 minutes, hour hand moves by 20°/2 = 10°
∴ Angle between hour & minute hand = 60° – 10° = 50°
(vi) 10:10
We know, angle between two clock mark is 30°
∴ Angle between hour & minute hand is less than (4×30°) = 120°
In one minute, hour hand moves by 1/2 o
In 10 minutes, hour hand moves by 10°/2 = 5°
∴ Angle between hour & minute hand is 120° – 5° = 115°
Exercise 1.2
(1) Find the length of an arc of a circle which subtends an angle of 108° at the centre, if the radius of the circle is 15 cm.
Solution:
Given, r = 15cm
θ = 108° = (108° × π/180°)C = 3πC/5
∴ Length of the arc, s = rθ = 15 × 3π/5 = 9π cm
(2) The radius of a circle is 9 cm. Find the length of an arc of this circle which cuts off a chord of length, equal to length of radius.
Solution:
Given, r = 9cm
∵ The chord length is equal to radius
∴ The triangle is an equilateral triangle
∴ θ = (60° × π/180°)C = πC/3
∴ Length of the arc, s = 9 × π/3 = 3π cm
(3) Find the angle in degree subtended at the centre of a circle by an arc whose length is 15 cm, if the radius of the circle is 25 cm.
Solution:
Given, s = 15cm
r = 25cm
∴ S = r θ
=> θ = S/r = 15/25 = 3C/5 = 3/5 × 180°/π = (108/π)°
(4) A pendulum of length 14 cm oscillates through an angle of 18°. Find the length of its path.
Solution:
Given, r = 14cm
θ = 18° = (18° × π/180°)C = πC/10
∴ arc length, S = r θ = 14 × π/10 = 7π/5 cm
(5) Two arcs of the same lengths subtend angles of 60° and 75° at the centres of two circles. What is the ratio of radii of two circles?
Solution:
Let, the two arcs have lengths r, cm and r2 cm
∴ 60° = 60° × π/180° = (π/3)C and 75° = 75° × π/180° = 5πC/12
∴ Arc length of 1st arc/Arc length of 2nd arc = (r1 π/3)/(r2 5π/212)
=> S1/s2 = 4r1/5r2 => r1/r2 = 5/4 (∵ S1 = S2)
(6) The area of a circle is 25π sq.cm. Find the length of its arc subtending an angle of 144° at the centre. Also find the area of the corresponding sector.
Solution:
Given, area of a circle = 25π cm2
=) πr2 = 25π
=) r2 = 25
=) r = 5cm
and, θ = 144° = 144 × π/180 = 4πC/5
∴ Arc length, s = r θ
= 5 × 4π/5
= 4π cm
Area of the sector, A = 1/2 π2 θ
= 1/2 × 5 × 5 x 4π/5
= 10π cm2
(7) OAB is a sector of the circle having centre at O and radius 12 cm. If m ∠AOB = 45°, find the difference between the area of sector OAB and triangle AOB.
Solution:
Given, r = 12cm
θ = 45° = 45° × π/180° = πC/4
Area of sector = 1/2 r2 θ = 1/2 × 12 × π/4
= 18π cm2
In △OAB, m ∠AOB = 45°
OA = OB = 12cm
Let, AD⊥OB
In △ADO, sin 45° = AD/OA
=) AD = 12 × 1/√2 = 6√2 cm
Area of △AOB = ½ × 12 × 6√2
= 36√2 cm2
∴ Difference = (18π – 36√2) cm2
= 18 (π – 2√2) cm2
(8) OPQ is the sector of a circle having centre at O and radius 15 cm. If m ∠POQ = 30°, find the area enclosed by arc PQ and chord PQ.
Solution:
Given, r = 15 cm
θ = 30° = 30° × π/180° = πC/6
Area of sector, A = 1/2 r2 θ = ½ × 15 × π/6
= 75π/4 cm2
In △OPQ, m ∠POQ = 30°
OP = OQ = 15cm
Let, PR⊥OQ,
In OPR, sin 30° = PR/OP => PR = 15×1/2 = 7.5 cm
Area of △OPQ = 1/2 × 15 × 7.5 = 56.25 cm2 = 225/4 cm2
∴ Require are = (75π/4 – 225/4) cm2 = 75/4 (π – 3) cm2
(9) The perimeter of a sector of the circle of area 25π sq.cm is 20 cm. Find the area of the sector.
Solution:
Let, r be radius of the circle
For sector OAB of circle centred at 0.
Perimeter = 20 cm
= l (OA) + l(OB) + l (arc AB) = 20
= r + r + rθ = 20
= r (2 + θ) =20 —– (i)
Given, area of circle = πr2
=) 25π = πr2
=) r = 5cm
∴ (i) = 5 (2 + θ) = 20
= 2 + θ = 4
= θ = 2C
∴ Area of sector = ½ r2 θ = ½
= 25 cm2
(Q10) The perimeter of the sector of the circle of area 64π sq.cm is 56 cm. Find the area of the sector.
Solution:
Let, r for radius of the circle
∴ Area of circle = πr2
= 64 π = πr2
= r = 8cm
For sector OAB circle centred at 0 perimeter = 56 cm
= l (OA) + l(OB) + l(arc AB) = 56
= 8 + 8 + 8 × θ = 56
= 8 (2 + θ) = 56
= 2 + θ = 7
= θ = 5C
Area of water, OAB = 1/2 r2 θ = ½ × 8 × 8 × 5 = 160 cm2
MISCELLANEOUS EXERCISE – 1
(I) Select the correct option from the given alternatives.
(1) (22π/15)C is equal to
(A) 246°
(B) 264°
(C) 224°
(D) 426°
Solution:
(22π/15)C
22π/15 × 180°/π = 264°
A) (B) 264°
(2) 156° is equal to
(A) (17π/15)C
(B) (13π/15)C
(C) (11π/15)C
(D) (7π/15)C
Solution:
156°
156° × π/180° = (13π/15)C
A) (B)
(3) 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 meters when it traces the angle of 72° at the centre, then the length of the rope is
(A) 70m
(B) 55m
(C) 40m
(D) 35m
Solution:
Given, S = 88m
θ = 72° = 72° × π/180° = 2πC/5
∴ S = r θ
=) r = S/θ = 88/(2π/5) = 220/π = 70m
A) (A) 70m
(4) If a 14cm long pendulum oscillates through an angle of 12°, then find the length of its path.
(A) 13π/14
(B) 14π/13
(C) 15π/14
(D) 14π/15
Solution:
Given, r = 14cm
θ = 12° = 12° × π/180° = πC/15
∴ Arc length, S = r θ = 14×π/15 = 14π/15 cm
(5) Angle between hands of a clock when it shows the time 9.45 is
(A) (7.5)°
(B) (12.5)°
(C) (17.5)°
(D) (22.5)°
Solution:
Angle between clock mark is 30°
But, angle between hour & minute hand is more than 0°
In one minute, hour hand moves by 1°/2
In 45 minutes, hour hand moves by 45°/2 = 22.5°
∴ Angle between hour & minute hand is 22.5°
A) (D) 22.5°
(6) 20 meters of wire is available for fancing off a flower-bed in the form of a circular sector of radius 5 meters, then the maximum area (in sq. m.) of the flower-bed is
(A) 15
(B) 22
(C) 25
(D) 30
Solution:
Given, A/Q Perimeter = 20m
= r + r + r θ = 20
= 5 (2 + θ) = 20
= 2 + θ = 4 = θ = 2C
∴ Maximum area, A = 1/2 r2 θ
= ½ × 5 × 5 × 2
= 25m2
A) (C) 25
(7) If the angles of a triangle are in the ratio 1:2:3, then the smallest angle in radian is
(A) π/3
(B) π/6
(C) π/2
(D) π/9
Solution:
Let, the angles be k, 2k & 3k
∴ K + 2K + 3K = 180°
= K = 30°
∴ Smallest angle = 30° = 30° × π/180° = πC/6
A: (B) π/6
(8) A semicircle is divided into two sectors whose angles are in the ratio 4:5. Find the ratio of their areas?
(A) 5:1
(B) 4:5
(C) 5:4
(D) 3:4
Solution:
Let, the angles be 4K and 5k
∴ 4K + 5K = 180° = 20°
∴ The angles are: 80° = 80° × π/180° = 4πC/9
100° = 100° × π/180° = 5πC/9
∴ Ratio of areas = A1/A2 = (1/2 π × 4π/9)/(1/2 r × 5π/9) = 4/5
∴ (B) 4:5
(9) Find the measure of the angle between hour-hand and the minute hand of a clock at twenty minutes past two.
(A) 50°
(B) 60°
(C) 54°
(D) 65°
Solution:
Angle between two clock mark is 30°
∴ Angle between hour and minute hand is less than (2×30°) = 60°
In one minute, hour hand moves by 1°/2
In 20 minutes, hour hand moves by 20°/2 = 10°
∴ Angle required = 60° – 10° = 50°
A: (A) 50°
(10) The central angle of a sector of circle of area 9π sq.cm is 60°, the perimeter of the sector is
(A) π
(B) 3 + π
(C) 6 + π
(D) 6
Solution:
Given,
θ = 60° = 60° × π/180° = πC/3
Area of circle, A = 9π
=) 9π = πr2 = 9π
=) r2 = 9
=) r = 3 cm
∴ Perimeter = r + r + r θ = r (2 + θ) = 3 (2 + π/3)
= 6 + π
A: (C) 6 + π
(II) Answer the following.
(1) Find the number of sides of a regular polygon if each of its interior angle is
3πC/4
Solution:
Given, interior angle = 3πC/4 = 3π/4 × 180/π = 135°
∴ Exterior angle = 180° – 135°
= 45°
∴ (360/n) = 45°
=) n = 360/45 = 8
(2) Two circles, each of radius 7 cm, intersect each other. The distance between their centres is 7 2 cm. Find the area of the portion common to both the circles.
Solution:
Let, the two circles interest each other at A and B. The centres are O & O′.
A/Q, OA = OA′ = OB = OB = 7cm
and OO′ = 7√2 cm
∴ OO′2 = 98
In △OAO′, OA2 + OA′2 = 72 + 72 = 98 = OO′2
∴ ∠OAO′ = ∠OBO′ = 90°
∴ OAO′B is a square
∴ ∠AOB = ∠AO′B = 90° = 90° × π/180° = πC/2
Now, in OAO′,
Area = 72 = 49 cm2
Area of sector OAB and O′ AB,
Area = 1/2 r2 θ
= 1/2 × 7 × 7 × π/2
= 49π/4 cm2
∴ Required area = area of sector OAB + Area of sector O′AB – area of square OAO′B
= 49π/4 + 49π/4 – 49
= 49 (π/2 – 1) cm2
(3) △PQR is an equilateral triangle with side 18 cm. A circle is drawn on the segment QR as diameter. Find the length of the arc of this circle within the triangle.
Solution:
Let, O be centre of the circle, the triangle intersects circle at A and B and QR is diameter.
Now, OA = OP = OB = OQ radius of circle = 18/2 = 9cm
Now, ∠OAP = ∠QRP = 60° (PQR is equilateral triangle)
Also, OA = OP = ∠OAP = ∠APO = 60°
∠AOQ = ∠OAP + ∠APO = 60° + 60° = 120°
and ∠AOP = 60°
Similarly, ∠BOQ = 60°
θ = ∠AOB = <AOQ – ∠BOQ = 120° – 60° = 60° = πC/3
∴ Arc length, S = r θ = 9 × π/3 = 3π cm
(4) Find the radius of the circle in which central angle of 60° intercepts an arc of length 37.4 cm.
Solution:
Given, θ = 60° = πC/3 and s = 37.4 cm
∴ S = r θ
= r = S/θ
= 37.4/(π/3)
= 37.4 × 3 × 7/22
= 36.7 cm
(5) 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?
Solution:
Given, r = 4cm
S = 10cm
∴ S = r θ
= 4θ = 10
= θ = 5C/2
= 5/2 × 180/π
= (450/π)°
(6) If two arcs of the same length in two circles subtend angles 65° and 110° at the centre. Find the ratio of their radii.
Solution:
Given, θ1 = 65° = 65° × π/180° = 13πC/36
θ2 = 110° = 110° × π/180° = 11πC/18
∴ ratio of arc length, S1/S2 = 1
= r1 θ1/r2 θ2 = 1
= (r1 13π/36)/(r2 11π/18) = 1
= r1/r2 = 11π/18 × 36/13π
= 22/13
(7) The area of a circle is 81π sq.cm. Find the length of the arc subtending an angle of 300° at the centre and also the area of the corresponding sector.
Solution:
Given, Area of circle = 81π cm2
= πr2 = 81π
= r = 9cm
θ = 300° = 300 × π/180 = 5πC/3
∴ Arc length, S = r θ = 9 × 5π/3 = 15π cm
Area of sector, A = 1/2 r2 θ
= ½ × 9 × 9 × 5π/3
= 135π/2 cm2
(8) Show that minute hand of a clock gains 5° 30′ on the hour hand in one minute.
Solution:
In one minute, hour hand makes 1°/2
In one minute, minute hand makes 360°/60° = 6°
∴ Gains of minute hand = 6° – 1/2° = 5.5° = 5° + (0.5 × 60)′
= 5° 30′
(9) 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.
Solution:
In one hour, distance covered = 36km
∴ In 30 seconds, distance covered = 36/60 × 1/2
= 3/10 km
∴ S = 3/10 km
= r θ = 3/10
= θ = (3/10)C (∵ r = 1 km)
∴ θ = 3/10 × 180/π = 54×7/ 22 = 17.18° = 17° + (0.18×60)′
= 17° 11′ (Approx)
(10) In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Solution:
Let, AB be the chord and O be the centre of the circle
∵ OA = OB = 20 cm (∵ diameter = 40cm)
(11) The angles of a quadrilateral are in A.P. and the greatest angle is double the least. Find angles of the quadrilateral in radian.
Solution:
Let, the angles be a – d, a, a + d, a + 2d
A/Q, a + 2d = 2 (a – d)
=) a = 4d —– (1)
a – d + a + a + d + a + 2d = 360°
= 4a + 2d = 360°
= 16d + 2d = 360°
= d = 360°/18 = 20°
∴ a = 4×20° = 80°
∴ The angles are, 60° = 60° × π/180° = πC/3
80° = 80° × π/180° = 4πC/9
100° = 100° × π/180° = 5πC/9
120° = 120° × π/180° = 2πC/3