RD Sharma Chapter 16 Class 9 Maths Exercise 16.4 Solutions

RD Sharma Chapter 16 Class 9 Maths Exercise 16.4 Solutions is one of the prime sections of the chapter called ‘Circle.’ Upon exercising the questions in these sections, the students can get in-depth knowledge about arcs formed upon these. The key to answering the questions based on these problems is always about following the steps right. Rather than traditional ways, a step-by-step method of answering the questions on it is a better approach. Otherwise, it might look complicated. Also, step by answering the questions can help a student have a better knowledge of the topic.

Students will get the varieties of questions in the attached PDF based on the RD Sharma Chapter 16 Class 9 Maths Exercise 16.4 Solutions. The explanation given in the PDF is prepared with easy solutions as per the learners’ understanding. All the formulas are mentioned with the proof along with the questions.

Learn about RD Sharma Chapter 16 Circles Class 9

Download RD Sharma Chapter 16 Class 9 Maths Exercise 16.4 Solutions PDF

Solutions for Class 9 Maths Chapter 16 Circles Exercise 16.4

Important Definitions RD Sharma Chapter 16 Class 9 Maths Exercise 16.4 Solutions

An angle subtended by the arc of a circle at its center is twice (two times) the angle it subtends anyplace on the circle’s circumference.

To prove the arcs and their subtended angles, we have to take the exterior angle theorem, where an exterior angle of a triangle is equivalent to the total of the opposite interior angles.

Examples of Arcs and Its Subtented Angles of RD Sharma Chapter 16 Class 9 Maths Exercise 16.4 Solutions

Ques- In the below figure, O is the center of a circle. If ∠APB = 50°, find ∠AOB and ∠OAB.

Solution-

= ∠APB = 50° (Given)

= By degree measure theorem: ∠AOB = 2∠APB

= ∠AOB = 2 × 50° = 100°

= Again, OA = OB [Radius of circle]

= Then ∠OAB = ∠OBA [Angles opposite to equal sides]

= Let ∠OAB = m

= In ΔOAB,

= By an angle sum property: ∠OAB+∠OBA+∠AOB=1800

= m + m + 100° = 180°

= 2m = 180° – 100° = 80°

= m = 80°/2 = 40°

=∠OAB = ∠OBA = 40°

Que- In the below figure, it is given that O is the centre of a circle and ∠AOC = 150°. Find ∠ABC.

Solution-

= ∠AOC = 150° (Given)

= By degree measure theorem-

= ∠ABC = (reflex ∠AOC)/2 …(1)

= We know, ∠AOC + reflex(∠AOC) = 360° [Complex angle]

= 150° + reflex ∠AOC = 360°

= or reflex ∠AOC = 360°−150° = 210°

=From (1) => ∠ABC = 210° /2 = 105°

Ques- In the following figure, O is the centre of a circle. Find ∠BAC.

Solution-

Given: ∠AOB = 80° and ∠AOC = 110°

= Therefore, ∠AOB+ ∠AOC+ ∠BOC=360° [Completeangle]

= Substitute of given values,

= 80°0 + 100° + ∠BOC = 360°

= ∠BOC = 360° – 80° – 110° = 170°

= or ∠BOC = 170°

= Now, by degree measure theorem

= ∠BOC = 2∠BAC

= 170° = 2∠BAC

= Or ∠BAC = 170°/2 = 85°

Frequently Asked Questions (FAQs) of RD Sharma Chapter 16 Class 9 Maths Exercise 16.4 Solutions

Ques 1- What is the angle subtended by an arc?

Ans- An angle is subtended by an arc, line segment, or some other section of a curve when its two rays cross through the endpoints of the arc, line segment, or curve section.

Ques 2- What is a subtended angle at the circumference?

Ans- If the endpoints of an arc are connected to a third point on the circumference of the circle, then an angle is created.

Ques 3- How do you find the length of an arc without an angle?

Ans- To measure arc length without an angle, we need the radius and the sector area (Multiply the area by 2). Then divide the outcome by the radius squared (ensure that the units are equal) to make the central angle in radians.

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