Maharashtra Board 11th Maths Chapter 1, Class 11 Maths Chapter 1 solutions
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.
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:
From the figure, the given angle terminates in quadrant III.
ii.
From the figure, the given angle terminates in quadrant III.
iii.
From the figure, the given angle terminates in quadrant I.
iv.
From the figure, the given angle terminates in quadrant I.
v.
From the figure, the given angle terminates in quadrant III.
vi.
From the figure, the given angle terminates in quadrant I.
vii.
From the figure, the given angle terminates in quadrant IV.
viii.
From the figure, the given angle terminates in quadrant I.
ix.
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.
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 = θ∘=(θ×π180)c
i. 85° = (85×π180)c=(17π36)c
ii. 250° = (250×π180)c=(25π18)c
iii. 132° = (−132×π180)c=(−11π15)c
[Note : Answer given in the textbook is 11π15 However, as per our calculation it is (−11π15)c ]
iv. 65° 30′ = 65° + 30′
= 65° + (3060)∘ … [1′ = (1/60)°]
= 65° + (1/2)°
v. 75° 30′ = 75° + 30′
vi. 40°48′ = 40° + 48′
Question 3.
Convert the following angles in degrees.
i. 7πc12
ii. −5πc3
iii. 5c
iv. 11πc18
v. (−14)c
Solution:
Question 4.
Express the following angles in degrees, minutes and seconds.
i. (183.7)°
ii. (245.33)°
iii. (15)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 180π)°
= (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.)
3k = 3 x 10° = 30°,
7k = 7 x 10° = 70° and
8k = 8 x 10° = 80°.
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)]
∴ ∠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 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°
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°
Interior angle + Exterior angle = 180°
∴ Each interior angle = 180° – 60° = 120°
iii. Septagon:
Number of sides = 7
Number of exterior angles = 7
Sum of exterior angles = 360°
Ex 1.2
Question 4.
A pendulum of length 14 cm oscillates through an angle of 18°. Find the length of its path.
Solution:
Draw AM ⊥ OB
In ΔOAM,
[Note: The question has been modified.]
Question 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:
Here, r = 15 cm
m∠POQ = 30°
\(\left(30 \times \frac{\pi}{180}\right)^{c}[/larex]
∴ θ =
Draw QM ⊥ OP
In ΔOQM,
