Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 1 Angle and its Measurement Ex 1.1 Questions and Answers.

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

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.

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

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:
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 1
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.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 3
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.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 5
From the figure, the given angle terminates in quadrant III.

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

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.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 7
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.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 9
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:
we know that = \(\theta^{\circ}=\left(\theta \times \frac{\pi}{180}\right)^{c}\)
i. 85° = \(\left(85 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{17 \pi}{36}\right)^{\mathrm{c}}\)
ii. 250° = \(\left(250 \times \frac{\pi}{180}\right)^{c}=\left(\frac{25 \pi}{18}\right)^{c}\)
iii. 132° = \(\left(-132 \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(-\frac{11 \pi}{15}\right)^{\mathrm{c}}\)
[Note : Answer given in the textbook is \(\frac{11 \pi}{15}\) However, as per our calculation it is \(\left(\frac{-11 \pi}{15}\right)^{c}\) ]

iv. 65° 30′ = 65° + 30′
= 65° + \(\left(\frac{30}{60}\right)^{\circ}\) … [1′ = (1/60)°]
= 65° + (1/2)°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 11

v. 75° 30′ = 75° + 30′
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 12

vi. 40°48′ = 40° + 48′
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 13

Question 3.
Convert the following angles in degrees.
i. \(\frac{7 \pi^{c}}{12}\)
ii. \(\frac{-5 \pi^{c}}{3}\)
iii. 5c
iv. \(\frac{11 \pi^{c}}{18}\)
v. \(\left(\frac{-1}{4}\right)^{c}\)
Solution:
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 14
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 15

Question 4.
Express the following angles in degrees, minutes and seconds.
i. (183.7)°
ii. (245.33)°
iii. \(\left(\frac{1}{5}\right)^{c}\)
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″

iii. We know that θc = (θ x \(\frac{180}{\pi}\))°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 16
= (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.)

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

Question 5.
In △ABC, if m∠A = \(\frac{7 \pi^{\mathrm{c}}}{36}\), m∠B = 120°, find m∠C in degree and radian.
Solution:
We know that θ c = (θ x \(\left(\theta \times \frac{180}{\pi}\right)^{\circ}\) ) °
In △ABC,
m∠A = \(\frac{7 \pi^{\mathrm{c}}}{36}=\left(\frac{7 \pi}{36} \times \frac{180}{\pi}\right)^{\circ}\) = 35°
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°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 17
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 18

Question 6.
Two angles of a triangle are \(\frac{5 \pi}{9}^{\mathrm{c}}\) and \(\frac{5 \pi}{18}^{\mathrm{c}}\) Find the degree and radian measures of third angle.
Solution:
We know that θc = [θ x \( ]°
The measures of two angles of a triangle are [latex]\frac{5 \pi^{\mathrm{c}}}{9}, \frac{5 \pi^{\mathrm{c}}}{18},\)
i.e., \(\left(\frac{5 \pi}{9} \times \frac{180}{\pi}\right)^{\circ},\left(\frac{5 \pi}{18} \times \frac{180}{\pi}\right)^{0}\)
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°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 19
∴ The degree and radian measures of the third angle are 30° and \(\left(\frac{\pi}{6}\right)^{\mathrm{c}}\) respectively.

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°.
we known that θ° = ( θ x \(\frac{\pi}{180}\)) c
∴ The measure of the angles in radius are
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 20

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
x + y = 5πc
x + y = (5π x \( \frac{180}{\pi}\) ) …[ θc = \(\left(\theta \times \frac{180}{\pi}\right)^{\circ}\) ]
∴ 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°.

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

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)]
∴ d = \(\frac{120^{\circ}}{3}\) = 40°
∴ 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.
In a cyclic quadrilateral two adjacent angles are 40 and \(\frac{\pi^{c}}{3}\). Find the angles of the quadralateral in degrees.
Solution:
Let ABCD be the cyclic quadrilateral such that
∠A = 40° and
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 23
∴  ∠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°.

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

Question 12.
One angle of a quadrilateral has measure \(\frac{2 \pi^{c}}{5}\) and the measures of other three angles are in the ratio 2:3:4. Find their measures in degrees and radians.
Solution:
We know that θc = \(\left(\theta \times \frac{180}{\pi}\right)^{\circ}\))
One angle of the quadrilateral has measure\(\frac{2 \pi^{c}}{5}=\left(\frac{2 \pi}{5} \times \frac{180}{\pi}\right)^{\circ}=72^{\circ}\)
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°
We know that θ° = (θ x \(\frac{\pi}{180}\))c
∴ The measures of the angles in radians are
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 24

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°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 25
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)°
= \(\left(\frac{360}{7} \times \frac{\pi}{180}\right)^{\mathrm{c}}=\left(\frac{2 \pi}{7}\right)^{\mathrm{c}}\)
Interior angle + Exterior angle = 180°
∴ Each interior angle = 180° – ( \(\frac{360}{7}\))°
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 27

iv. Octagon:
Number of sides = 8
Number of exterior angles = 8
Sum of exterior angles = 360°
∴ Each exterior angle = \(\frac{360^{\circ}}{\text { no. of sides }}=\frac{360^{\circ}}{8}\)
= 45°
= \(\left(45 \times \frac{\pi}{180}\right)^{c}=\left(\frac{\pi}{4}\right)^{c}\)
Interior angle + Exterior angle = 180°
Each interior angle = 180° – 45° = 135°
= \(\left(135 \times \frac{\pi}{180}\right)^{c}=\left(\frac{3 \pi}{4}\right)^{c}\)

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 \(\left(\frac{1}{6}\right)^{\text {th }}\) of the angle between 11 and 12.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 28
Angle between two consecutive marks = \(\frac{360^{\circ}}{12}\) = 30°
Angle traced by hour-hand in 10 minutes
= \(\frac{1}{6}\) (30°) = 5°
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 \(\left(\frac{1}{3}\right)^{ }\)rd of angle between 7 and 8.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 29
Angle between two consecutive marks
= 360°/12 = 30°
Angle traced by hour-hand in 20 minutes
= \(\frac{1}{3}\)(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 \(\left(\frac{7}{12}\right)^{ }\)th of angle between 1 and 2.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 30
Angle between two consecutive marks
= 360°/12 = 30°
Angle traced by hour-hand in 35 minutes
= \(\frac{7}{12}\left(30^{\circ}\right)=\left(\frac{35}{2}\right)^{\circ}=\left(17 \frac{1}{2}\right)^{\circ}\frac{1}{3}\)
Angle between marks 1 and 7 = 6 x 30° = 180°
Angle between two hands of the clock at thirty five past one = 180° – \(\left(17 \frac{1}{2}\right)^{\circ}=\left(162 \frac{1}{2}\right)^{\circ}\)
= 162° + \(\frac{1}{2}\) = 162°30′

iv. At 5:45, the minute-hand is at mark 9 and hour- hand has crossed ( \(frac{3}{4}\) )th of the angle between 5 and 6.
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 31
Angle between two consecutive marks
= 360°/12 = 30°
Angle traced by hour-hand in 45 minutes
\(\frac{3}{4}\left(30^{\circ}\right)=(22.5)^{\circ}=\left(22 \frac{1}{2}\right)^{\circ}\)
Angle between marks 5 and 9
= 4 x 30° = 120°
∴ Angle between two hands of the clock at quarter to six = \(120^{\circ}-\left(22 \frac{1}{2}\right)^{0}\)
Maharashtra Board 11th Maths Solutions Chapter 1 Angle and its Measurement Ex 1.1 32

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

v. At 2 : 20, the minute-hand is at mark 4 hour hand has crossed \(\frac{1}{3}\)rd 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
= \(frac{1}{3}\)(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
= \(\frac{1}{6}\) (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°

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