NCERT Class 9 Solutions: Circles (Chapter 10) Exercise 10.5 – Part 2

Exterior angles on circle- 3 types of exteriro angle formed by (1) two secants (2) a secant and a tangent (3) 2 tangents

Exterior Angles on Circle

Exterior angles on circle- 3 types of exteriro angle formed by (1) two secants (2) a secant and a tangent (3) 2 tangents

Q-3 In the figure ,ABC=100° , where A, B and C are points on a circle with centre O. Find OAC.

Circle c Circle c: Circle through C with center O Angle ? Angle ?: Angle between A, B, A' Angle ? Angle ?: Angle between A, B, A' Angle ? Angle ?: Angle between A, B, A' Segment f Segment f: Segment [C, A] Segment g Segment g: Segment [A, B] Segment h Segment h: Segment [B, C] Segment i Segment i: Segment [A, O] Segment j Segment j: Segment [C, O] Point O O = (0.16, 2.66) Point O O = (0.16, 2.66) Point O O = (0.16, 2.66) Point C C = (2.58, 3.06) Point C C = (2.58, 3.06) Point C C = (2.58, 3.06) Point A Point A: Point on c Point A Point A: Point on c Point A Point A: Point on c Point B Point B: Point on c Point B Point B: Point on c Point B Point B: Point on c

a, B, C Are Points on a Circle With Center O

A, B, C are points on a circle with center O, ∠ABC = 100°

Solution:

  • Given A, B, and C are points on a circle with center O

  • ABC=100°

The major arc AC subtends an angle AOC at the center and an angle ABC on rest of the circle. Therefore,

AOC=2×ABC=2×100°=200°

AOC=360°200°=160° (the angle inside the quadrilateral ABCO).

In ΔOAC , OA=OC (radius of the circle), therefore the triangle is isosceles and hence, OAC=OCA

Now,

  • OAC+OCA+AOC=180° (Sum of the angles in triangle)

  • OAC+OAC+160°=180° ( OAC=OCA )

  • 2OAC=180°160°

  • OAC=10°

Q-4 In the figure, PQR=69° , ∠ PRQ=31° , find QSR.

Circle c Circle c: Circle through R with center A Angle ? Angle ?: Angle between P, Q, R Angle ? Angle ?: Angle between P, Q, R Angle ? Angle ?: Angle between P, Q, R Angle ? Angle ?: Angle between P, R, Q Angle ? Angle ?: Angle between P, R, Q Angle ? Angle ?: Angle between P, R, Q Segment f Segment f: Segment [Q, R] Segment g Segment g: Segment [Q, S] Segment h Segment h: Segment [S, R] Segment i Segment i: Segment [R, P] Segment j Segment j: Segment [P, Q] Point R R = (2.52, 3) Point R R = (2.52, 3) Point R R = (2.52, 3) Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point S Point S: Point on c Point S Point S: Point on c Point S Point S: Point on c Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c

Points P, Q, R, S on Circle

P, Q, R, S are on circle, ∠ABC = 69°, ∠ ACB = 31°

Solution:

Given

  • P, Q, R, and S are points on a the circle

  • PQR=69° , PRQ=31°

Note that, QPR=QSR (Angles subtended by the same segment on the circle)

In ΔPQR,

  • QPR+PQR+PRQ=180° (Sum of the angles on a triangle)

  • QPR+69°+31°=180° (PQR=69° , ∠ PRQ=31°)

  • QPR=180°100°

  • QPR=80°

Thus, QSR=80°

Q-5 In the figure, P, Q, R and S are four points on a circle. PQ and SR intersect at a point E such that ∠SEQ = 130° and ∠EQR = 20°. Find ∠ SPQ.

Circle c Circle c: Circle through B with center A Angle ? Angle ?: Angle between S, E, C' Angle ? Angle ?: Angle between S, E, C' Angle ? Angle ?: Angle between S, E, C' Angle ? Angle ?: Angle between R, Q, F' Angle ? Angle ?: Angle between R, Q, F' Angle ? Angle ?: Angle between R, Q, F' Segment f Segment f: Segment [S, Q] Segment g Segment g: Segment [S, E] Segment h Segment h: Segment [E, Q] Segment i Segment i: Segment [R, Q] Segment j Segment j: Segment [R, S] Segment k Segment k: Segment [P, E] Segment l Segment l: Segment [P, S] Point S Point S: Point on c Point S Point S: Point on c Point S Point S: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point Q Point Q: Point on c Point E E = (-0.11, 5.34) Point E E = (-0.11, 5.34) Point E E = (-0.11, 5.34) Point R Point R: Point on c Point R Point R: Point on c Point R Point R: Point on c Point P Point P: Point on c Point P Point P: Point on c Point P Point P: Point on c

P, Q, R, S Four Points on a Circle.

P, Q, R, S four points on a circle .PQ and SR intersect at a point E,∠SEQ = 130° and ∠EQR = 20°.

Solution:

Given,

  • P, Q, R, S are four points on a circle.

  • PQ and SR intersect at a point E

  • SPQ=QRE (Angles in the segment of the circle)

In ΔQRE ,

  • QES=QRE+RQE (Exterior angles of the triangle)

  • 130°=QRE+20°

  • QRE=110°

So, SPQ=110° (SPQ=QRE ) (Angles subtended by the same arc)

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