Chapter 1 Angle and its Measurement Miscellaneous Exercise 1 Solutions

(B) 264°

(B)

(A) 70 m

(D)

(D) (22.5)°

(C) 25 r + r + rθ = 20m 2r + rθ = 20

(B)

(B) 4:5

(A) 50°

(C) 6 + π

Each interior angle of a regular polygon = = 135° Interior angle + Exterior angle = 180° ∴ Exterior angle = 180° – 135° = 45° Let the number of sides of the regular polygon be n. But in a regular polygon, exterior angle = ∴ 45° = ∴ n = = 8 ∴ Number of sides of a regular polygon = 8.

Let O and O1 be the centres of two circles intersecting each other at A and B. Then OA = OB = O1A = O1B = 7 cm and OO1 = 7√2 cm OO12 = 98 ………………(i) Since OA2 + O1A2 = 72 = 98 = OO12 …..[ from (i)] m∠OAO1 = 90° □ OAO1B is a square. m∠AOB = m∠AO1B = 90° A(□ OAO1B) = (side)2 = (7)2 = 49 sq.cm ∴ Required area = area of shaded portion = A(sector OAB) + A(sector O1AB)) – A(□ OAO1B)

Let ‘O’ be the centre of the circle drawn on QR as a diameter. Let the circle intersect seg PQ and seg PR at points M and N respectively. Since l(OQ) = l(OM), m∠OM Q = m∠OQM = 60° m∠MOQ = 60° Similarly, m∠NOR = 60° Given, QR =18 cm. r = 9 cm θ = 60° = (60 x )c = ∴ l(arc MN) = S = rθ = 9 x = 3π cm.

Let S be the length of the arc and r be the radius of the circle. θ = 60° = S = 37.4 cm Since S = rθ,

S = 10 cm and r = 4 cm Since S = rθ, 10 = 4 x θ

Let r1 and r2 be the radii of the two circles and let their arcs of same length S subtend angles of 65° and 110° at their centres. Angle subtended at the centre of the first circle, Angle subtended at the centre of the second circle,

Area of circle = πr2 But area is given to be 81 n sq.cm ∴ πr2 = 81π ∴ r2 = 81 ∴ r = 9 cm θ = 300° = Since S = rθ S = 9 x = 15π cm Area of sector = x r x S = x 9 x 15π = sq.cm

Angle made by hour-hand in one minute Angle made by minute-hand in one minute = = 6° ∴ Gain by minute-hand on the hour-hand in one minute = = 5°30′ [Note: The question has been modified.]

r = 1km = 1000m l(Arc covered by train in 30 seconds) = 30 x m ∴ S = 300 m Since S = rθ, 300 = 1000 x θ = (17.18)° = 17° +(0.18)° = 17° + (0.18 x 60)’ = 17° + (10.8)’ ∴ θ = 17°11′(approx.)

Let ‘O’ be the centre of the circle and AB be the chord of the circle. Here, d = 40 cm ∴ r = = 20 cm Since OA = OB = AB, ∆OAB is an equilateral triangle. The angle subtended at the centre by the minor arc AOB is θ = 60° = = l(minor arc of chord AB) = rθ = 20 x = cm

Let the measures of the angles of the quadrilateral in degrees be a – 3d, a – d, a + d, a + 3d, where a > d > 0 ∴ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 360° … [Sum of the angles of a quadrilateral is 360°] ∴ 4a = 360° ∴ a = 90° According to the given condition, the greatest angle is double the least, ∴ a + 3d = 2.(a – 3d) ∴ 90° + 3d = 2.(90° – 3d) ∴ 90° + 3d = 180° – 6d 9d = 90° ∴ d = 10° ∴ The measures of the angles in degrees are a – 3d = 90° – 3(10°) = 90° – 30° = 60°, a – d = 90° – 10° = 80°, a + d = 90°+ 10°= 100°, a + 3d = 90° + 3(10°) = 90° + 30° = 120°