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Learn/ 10th Mathematics/ Unit 9 /Numericals

Unit 9: Tangent and Angles — Exercise 9 2

10th Class Mathematics · Unit 9: Tangent and Angles

9.2.1.A, B, C and D are points on the circumference of a circle with centre at O as shown in figure (B, D, C lie such that C is on ray BD extended, and the exterior angle at C is 114°). Find the angle x, the central angle subtending arc BD.
Given
Exterior angle Exterior angle at C = 114^circ
Formula
Exterior angle theorem 114^circ = angle BAD + angle ABD
Angles in same segment angle BAD = angle BCD and angle ABD = angle BCD (both stand on arc BD)
Let angle BAD = angle ABD = angle BCD = y ⇒ 114^circ = 2y
y = 57^circ
Central angle theorem x = 2y
Result
Value of x x = 2 × 57^circ = 114^circ
9.2.2.In the adjoining figure m∠BDC = 74°, find m∠BAC, m∠BOC and m∠OBC.
Given
Given angle angle BDC = 74^circ
Formula
Central angle theorem angle BOC = 2 × angle BDC
Working
Substituting angle BOC = 2 × 74^circ = 148^circ
Formula
Isosceles \triangle OBC (OB = OC) angle OBC = angle OCB = x
Working
Angle sum x + x + 148^circ = 180^circ
2x = 32^circ ⇒ x = 16^circ
Formula
Inscribed angle theorem angle BAC = 12 × angle BOC = 12 × 148^circ
Result
Answers angle BOC = 148^circ, angle OBC = 16^circ, angle BAC = 74^circ
9.2.3.Find the angles x and y in the following figures. (i) ABCD is a cyclic quadrilateral with ∠A = 74° and ∠B = 159°; x = ∠D, y = ∠C. (ii) O is the centre of a circle, P, Q, R lie on the circle with OP = OQ and ∠OPQ = ∠OQP = 38°; x = ∠PRQ (inscribed angle on arc PQ), y = ∠OQP. (iii) O is the centre of a circle, X and Y lie on the circle, OY is a radius perpendicular to the tangent at Y, and ∠XOY = 62°; x = ∠OXY, y is the angle formed with the tangent line at Y.
Given
(i) Given angles angle A = 74^circ, angle B = 159^circ
Formula
(i) Cyclic quadrilateral, opposite angles supplementary angle A + angle C = 180^circ ⇒ angle C = 106^circ
Working
(i) angle B + x = 180^circ ⇒ 159^circ + x = 180^circ
Result
(i) Result x = 21^circ, y = angle C = 106^circ
Given
(ii) Given OP = OQ (radii), angle OPQ = angle OQP = 38^circ
Formula
(ii) Angle sum of \triangle OPQ angle POQ = 180^circ - (38^circ + 38^circ) = 104^circ
(ii) Inscribed angle theorem angle PRQ = 12 × angle POQ = 12 × 104^circ
Result
(ii) Result x = 52^circ, y = 38^circ
Given
(iii) Given OY = OX (radii), angle XOY = 62^circ, angle OYX = 90^circ (radius perp tangent)
Formula
(iii) Angle sum of \triangle OXY x + 62^circ + 90^circ = 180^circ
x = 180^circ - 152^circ = 28^circ
y = 180^circ - (90^circ + 28^circ) = 62^circ
Result
(iii) Result x = 28^circ, y = 62^circ
9.2.4.Find the angle x in the given figure. (A, B, C, D are points on a circle with centre O; AB = AD (marked equal chords); rays from C form an exterior angle ∠ at C of 70°; x = ∠COD.)
Given
Given exterior angle Exterior angle at C = 70^circ
Formula
Exterior angle theorem 70^circ = angle CAB + angle CDA
Equal chords subtend equal angles at circumference AB = AD ⇒ angle CAB = angle CDA = y
70^circ = y + y = 2y ⇒ y = 35^circ
Central angle theorem angle COD = 2 × y
Result
Value of x x = 2 × 35^circ = 70^circ
9.2.5.In a circular arch over a monument entrance, two spotlight fixtures are placed such that they each shine from two different points on the arch to the same chord PQ on the base of the arch. If the angle formed at one light is 55°, what is the angle at the second light on the same side of chord PQ?
Given
Given angle Angle at first light = 55^circ
Formula
Angles on same chord, same side, are equal Let second angle = x
Result
Angle at second light x = 55^circ
9.2.6.A circular garden has a walking path forming a quadrilateral inscribed in it. If one angle is 87°, what is the opposite angle?
Given
Given angle One angle = 87^circ
Formula
Cyclic quadrilateral, opposite angles supplementary Opposite angle = 180^circ - 87^circ
Result
Opposite angle 93^circ
9.2.7.A ferris wheel has a radius of 12 m. If a passenger travels a distance of 18 m along the circumference of the ferris wheel, what is the angle (in radians) swept by the passenger's position from the starting point?
Given
Given ell = 18 m, r = 12 m
Formula
Arc length formula ell = rθ ⇒ θ = ellr
Working
Substituting θ = 1812 = 32
Result
Angle swept θ = 1.5 radians
9.2.8.Find θ, when (i) ℓ = 3 cm, r = 2.2 cm (ii) ℓ = 5.6 cm, r = 2 cm.
Formula
Formula ell = rθ ⇒ θ = ellr
Working
(i) θ = 32.2 = 3022 = 1511
Result
(i) Result θ ≈ 1.36 rad
Working
(ii) θ = 5.62
Result
(ii) Result θ = 2.8 rad
9.2.9.Find r, when (i) ℓ = 5.5 cm, θ = 40°20' (ii) ℓ = 13 cm, θ = 70°.
Formula
Formula ell = rθ ⇒ r = ellθ (θ in radians)
(i) Convert to radians θ = 40^circ20' = 40 + 2060 = 1213^circ = 1213 × pi180 = 121pi540 rad
Working
(i) r = frac{5.5}{121pi540 = 5.5 × 540121pi = 2970121pi
Result
(i) Result r ≈ 7.81 cm
(ii) Convert to radians θ = 70^circ = 70 × pi180 = 7pi18 rad
Working
(ii) r = frac{13}{7pi18 = 13 × 187pi = 2347pi
Result
(ii) Result r ≈ 10.64 cm
9.2.10.Find ℓ and area of sector, when (i) r = 1.7 cm, θ = 0.25 radian (ii) r = 3 cm, θ = 45°.
Given
(i) Given r = 1.7 cm, θ = 0.25 rad
Formula
(i) Arc length and sector area ell = rθ,quad Area = 12r2θ
Working
(i) ell = 1.7 × 0.25 = 0.425 cm
(i) Area = 12 × (1.7)2 × 0.25 = 12 × 2.89 × 0.25
Result
(i) Result ell = 0.425 cm, Area = 0.36125 cm2
Given
(ii) Given r = 3 cm, θ = 45^circ
(ii) Convert to radians θ = 45^circ = 45 × pi180 = pi4 rad
Working
(ii) ell = rθ = 3 × pi4 = 3pi4 cm
(ii) Area = 12 × 32 × pi4 = 9pi8 cm2
Result
(ii) Result ell = 3pi4 cm, Area = 9pi8 cm2
9.2.11.Uzma cut a pizza of radius 14 cm into 8 equal slices. What is the area of one slice (sector)?
Given
Given r = 14 cm, slices = 8
Formula
Total area of pizza pi r2 = pi (14)2 = 196pi cm2
Area of one slice frac{Total area{Number of slices = 196pi8
Result
Area of one slice 24.5pi cm2
9.2.12.The perimeter and area of a sector are 14 cm and 10 cm² respectively. Find the radius of the circle and the central angle of the sector.
Given
Given Perimeter = 2r + ell = 14,quad Area = 12rell = 10
From perimeter equation ell = 14 - 2r
Working
Substituting into area equation 12r(14 - 2r) = 10 ⇒ r(14-2r) = 20
14r - 2r2 = 20 ⇒ 2r2 - 14r + 20 = 0 ⇒ r2 - 7r + 10 = 0
Factoring (r-5)(r-2) = 0 ⇒ r = 5 cm or r = 2 cm
Checking both roots r=2 ⇒ ell = 10, Area = 10 checkmark; quad r=5 ⇒ ell = 4, Area = 10 checkmark
Result
Chosen radius r = 2 cm
Formula
Central angle ell = rθ ⇒ θ = ellr = 102
Result
Final answer Radius = 2 cm, Central angle = 5 rad