Chapter 1 Angle and its Measurement Ex 1.2 Solutions

Here, r = 15cm and θ = 108° = Since S = r.θ S = 15 x = 9π cm.

Here, r = 9cm Let the arc AB cut off a chord equal to the radius of the circle. Since OA = OB = AB, ΔOAB is an equilateral triangle. m∠AOB = 60° θ = 60° = Since S = r.θ, S = 9 x = 3π cm.

Here, r = 25 cm and S = 15 cm Since S = r.θ, 15 = 25 x θ ∴ The required angle in degree is or (34.40)°(approx.).

Let r1, and r2 be the radii of the two circles and let their arcs of same length S subtend angles of 60° and 75° at their centres. Angle subtended at the centre of the first circle, θ1 = 60° = ∴ S = r1θ1 = r1() Angle subtended at the centre of the second circle,

Area of circle = πr2 But area is given to be 25 π sq.cm ∴ 25π = πr2 ∴ r2 = 25 ∴ r = 5 cm θ = 144° = Since s = rθ S = 5() = 4π Also, A(sector) = x r x S = x 5 x 4π = 10π sq. cm

Here, r = 12 cm Draw AM ⊥ OB In ΔOAM, [Note: The question has been modified.]

Here, r = 15 cm m∠POQ = 30° Draw QM ⊥ OP In ΔOQM, sin 30° = QM= 15 x Shaded portion indicates the area enclosed by arc PQ and chord PQ. ∴ A(shaded portion) = A(sector OPQ) – A(ΔOPQ)

Area of circle = πr2 But area is given to be 25π sq.cm. ∴ 25π = πr2 ∴ r2 = 25 ∴ r = 5 cm Perimeter of sector = 2r + S But perimeter is given to be 20 cm. ∴ 20 = 2(5) + S ∴ 20 = 10 + S ∴ S = 10 cm Area of sector = x r x S = x 5 x 10 = 25sq.cm.

Area of circle = πr2 But area is given to be 25π sq.cm. ∴ 64π = πr2 ∴ r2 = 64 ∴ r = 8 cm Perimeter of sector = 2r + S But perimeter is given to be 20 cm. ∴ 56 = 2(5) + S ∴ 56 = 16 + S ∴ S = 40 cm Area of sector = x r x S = x 8 x 40 = 160sq.cm.