Maharashtra Board Class 11 Maths Part 1 Chapter 1 Angle and its Measurement Ex 1.1 Solution

Question 1.
(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:
210°,- 150°
210°-(- 150°) = 210°+ 150°
= 360°
= 1 (360°),
which is a multiple of 360°.
∴ The given pair of angles is co-terminal.

ii. 360°, – 30°
360° – (- 30°) = 360° + 30°
= 390°,
which is not a multiple of 360°.
∴ The given pair of angles is not co-terminal.

iii. -180°, 540°
540° -(-180°) = 540°+ 180°
= 720°
= 2(360°),
which is a multiple of 360°.
.’. The given pair of angles is co-terminal.

iv. – 405°, 675°
675° – (- 405°) = 675° + 405°
= 1080°
= 3(360°),
which is a multiple of 360°.
.’. The given pair of angles is co-terminal.

v. 860°, 580°
860° – 580° = 280°
which is not a multiple of 360, °.
∴ The given pair of angles is not co-terminal.

vi. 900°, 900°
900° – (-900°) = 900° + 900°
= 1800°
= 5(360°)
which is a multiple of 360°
∴ The given pair of angles is co-terminal.

Question 1.
(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:

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From the figure, the given angle terminates in quadrant III.

ii.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 2
From the figure, the given angle terminates in quadrant III.

iii.
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From the figure, the given angle terminates in quadrant I.

iv.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 4
From the figure, the given angle terminates in quadrant I.

v.
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From the figure, the given angle terminates in quadrant III.

vi.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 6
From the figure, the given angle terminates in quadrant I.

vii.
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From the figure, the given angle terminates in quadrant IV.

viii.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 8
From the figure, the given angle terminates in quadrant I.

ix.
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From the figure, the given angle terminates in quadrant III.
[Note: Answer given in the textbook is ‘Angle lies in quadrant II’. However, we found that it lies in quadrant III.]

x.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 10
From the figure, the given angle terminates in quadrant III.

Question 2.
Convert the following angles into radians,
i. 85°
ii. 250°
iii. -132°
iv. 65°30′
v. 75°30′
vi. 40°48′
Solution:
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v. 75° 30′ = 75° + 30′
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vi. 40°48′ = 40° + 48′
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Question 3.
Convert the following angles in degrees.

Solution:

Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 14
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Question 4.
Express the following angles in degrees, minutes and seconds.
i. (183.7)°
ii. (245.33)°

Solution:

We know that 1° = 60′ and 1′ = 60″
i. (183.7)° = 183° +(0.7)°
= 183° + (0.7 x 60)’
= 183°+ 42′
= 183° 42′

ii. (245.33)° = 245° + (0.33)°
= 245° + (0.33 x 60)’
= 245° + (19.8)’
= 245° + 19’+ (0.8)’
= 245° 19’+ (0.8 x 60)”
= 245° 19’+ 48″
= 245° 19′ 48″

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= (11.46)°
= 11° +(0.46)°
= 11° + (0.46×60)’
= 11°+ (27.6)’
= 11°+ 27’+ (0.6)’
= 11° + 27′ + (0.6×60)”
= 11°27′ + 36″
= 11°27’36” (approx.)

Question 5.
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m∠B = 120°
∴ m∠A + m∠B + m∠C = 180°
… [Sum of the angles of a triangle is 180°]
∴ 35° + 120° + m∠C = 180° m∠C = 180° – 35° – 120°
∴ m∠C = 25°
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Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 18

Question 6.
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i.e., 100°, 50°
Let the measure of third angle of the triangle be x°.
∴ 100°+50°+x° = 180°
…[Sum of the angles of a triangle is 180°]
∴ x° = 180°- 100° – 50°
∴ x° = 30°
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Question 7.
In a right angled triangle, the acute angles are in the ratio 4:5. Find the angles of the triangle in degrees and radians.
Solution:
Since the triangle is aright angled triangle, one of the angles is 90°.
In the right angled triangle, the acute angles are in the ratio 4:5.
Let the measures of the acute angles of the triangle in degrees be 4k and 5k, where k is a constant.
∴ 4k + 5k+ 90°= 180°
… [Sum of the angles of a triangle is 180°]
∴ 9k = 180° – 90°
∴ 9k = 90°
∴ k = 10°
∴ The measures of the angles in degrees are
4k = 4 x 10° = 40°,
5k = 5 x 10° = 50°
and 90°.
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∴ The measure of the angles in radius are
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Question 8.
The sum of two angles is 5πc and their difference is 60°. Find their measures in degrees.
Solution:

Let the measures of the two angles in degrees be x and y.
Sum of two angles is 5πc
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∴ x + y = 900° ………..(i)
∴ Difference of two angles is 60°.
x – y = 60° ….(ii)
Adding (i) and (ii), we get
2x = 960°
∴ x = 480°
Substituting the value of x in (i), we get
480° + y = 900°
∴ y = 900° — 480° = 420°
∴ The measures of the two angles in degrees are 480° and 420°.

Question 9.
The measures of the angles of a triangle are in the ratio 3:7:8. Find their measures in degrees and radians.
Solution:

The measures of the angles of the triangle are in the ratio 3:7:8.
Let the measures of the angles of the triangle in degrees be 3k, 7k and 8k, where k is a constant.
∴ 3k + 7k + 8k = 180°
… [Sum of the angles of a triangle is 180°]
∴ 18k =180°
∴ k = 10°
∴ The measures of the angles in degrees are
3k = 3 x 10° = 30°,
7k = 7 x 10° = 70° and
8k = 8 x 10° = 80°.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 21

Question 10.
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 degrees and radians.
Solution:

Let the measures of the angles of the triangle in degrees be a – d, a, a + d, where a > d> 0.
∴ a – d + a + a + d = 180°
…[Sum of the angles of a triangle is 180°]
∴ 3a = 180°
∴ a = 60° …(i)
According to the given condition, greatest angle is 5 times the smallest angle.
∴ a + d = 5 (a – d)
∴ a + d = 5a – 5d
∴ 6d = 4a
∴ 3d = 2a
∴ 3d = 2(60°) …[From (i)]
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∴ The measures of the angles in degrees are
a – d = 60° – 40° = 20°
a = 60° and
a + d = 60° + 40° = 100°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 22

Question 11.

Solution:

Let ABCD be the cyclic quadrilateral such that
∠A = 40° and
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∴  ∠A + ∠C = 180°
∴ 40° + ∠C = 180°
∴ ∠C= 180°- 40°= 140°
Also, ∠B + ∠D = 180°
… [Opposite angles of a cyclic quadrilateral are supplementary]
∴ 60° + ∠D =180°
∴ ∠D = 180°- 60° = 120°
∴ The angles of the quadrilateral in degrees are 40°, 60°, 140° and 120°.

Question 12.

Solution:
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Measures of other three angles are in the ratio 2:3:4.
Let the measures of the other three angles of the quadrilateral in degrees be 2k, 3k, 4k, where k is a constant.
∴ 72° + 2k + 3k + 4k = 360°
…[Sum of the angles of a quadrilateral is 360°]
∴ 9k = 288°
k = 32°
∴ The measures of the angles in degrees are
2k = 2 x 32° = 64°
3k = 3 x 32° = 96°
4k = 4 x 32°= 128°
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∴ The measures of the angles in radians are
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Question 13.
Find the degree and radian measures of exterior and interior angles of a regular
i. pentagon
ii. hexagon
iii. septagon
iv. octagon
Solution:

i. Pentagon:
Number of sides = 5
Number of exterior angles = 5
Sum of exterior angles = 360°
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Interior angle + Exterior angle = 180°
∴ Each interior angle = 180° — 72° = 108°
= \(

ii. Hexagon:
Number of sides = 6
Number of exterior angles = 6
Sum of exterior angles = 360°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 26
Interior angle + Exterior angle = 180°
∴ Each interior angle = 180° – 60° = 120°
= (120 x

[latex]\frac{\pi}{180}\))c = ( \(\frac{2 \pi}{3}[latex])c

iii. Septagon:

Number of sides = 7

Number of exterior angles = 7

Sum of exterior angles = 360°

∴ Each exterior angle = [latex]\frac{360^{\circ}}{\text { no. of sides }}=\frac{360^{\circ}}{7}\)

= (51.43)°

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iv. Octagon:

Number of sides = 8

Number of exterior angles = 8

Sum of exterior angles = 360°

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Question 14.
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. At 11:10, the minute-hand is at mark 2 and hour-hand has crossed   of the angle between 11 and 12.

Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 28

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Angle between marks 11 and 2 = 3 x 30° = 90°

∴ Angle between two hands of the clock at ten past eleven = 90° – 5° = 85°

ii. At 7 : 20 the minute -hand is at mark 4 and hour -hand has crossed ()rd of angle between 7 and 8.

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Angle between two consecutive marks

= 360°/12 = 30°

Angle traced by hour-hand in 20 minutes

=  (30°)= 10°

Angle between marks 4 and 7 = 3 x 30° = 90°

Angle between two hands of the clock at twenty past seven = 90° – 10° = 100°

iii. At 1 : 35 the minute -hand is at mark 7 and hour -hand has crossed ()th of angle between 1 and 2.

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Angle between two consecutive marks

= 360°/12 = 30°

Angle traced by hour-hand in 35 minutes

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Angle between marks 1 and 7 = 6 x 30° = 180°

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iv. At 5:45, the minute-hand is at mark 9 and hour- hand has crossed ( frac34 )th of the angle between 5 and 6.

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Angle between two consecutive marks

= 360°/12 = 30°

Angle traced by hour-hand in 45 minutes

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Angle between marks 5 and 9

= 4 x 30° = 120°

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v. At 2 : 20, the minute-hand is at mark 4 hour hand has crossed 1/3rd of the angle between 2 and 3.

Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 33

Angle between two consecutive marks = 360°/12 = 30°

Angle traced by hour-hand in 20 minutes

= frac13(30°)= 10°

Angle between marks 2 and 4 = 2 x 30° = 60°

∴ Angle between two hands of the clock at 2 :20 = 60° – 10° = 50°

.vi. At 10:10, the minute-hand is at mark 2 and hour-hand has crossed\frac{1}{6}[/latex]

th between 10 and 11.

Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 34

Angle between two consecutive marks
360°/12 = 30°
Angle traced by hour-hand in 10 minutes
=1/6 (30°) = 5°
Angle between marks 10 and 2= 4 x 30° = 120°
… Angle between two hands of the clock at 10:10
= 120° – 5°= 115°