Abcd is a cyclic quadrilateral such that ab is a diameter of the circle circumscribing it

Given: ABCD is a cyclic quadrilateral whose side AB is the diameter of the circle and ∠ A D C = 130 ∘. To find ∠ B A C . ∠ D + ∠ B = 180 ∘ (Opposite angles of a cyclic quadilateral ) 130 ∘ + ∠ B = 180 ∘ ∠ B = 180 ∘ − 130 ∘ = 50 ∘ ∠ A C B = 90 ∘ (Angle in a semicircle) In Δ A B C , ∠ B A C + 50 ∘ + 90 ∘ = 180 ∘ (Since sum of angles of a triangle is 180 ∘ ) ∠ B A C = 180 ∘ − 90 ∘ − 50 ∘ = 40 ∘

SOLUTION (B) 50 ∘ Given ABCD is a cyclic quadrilateral and ∠ A D C = 140 ∘ We know that, sum of the opposite angles in a cyclic quadrilateral is 180 ∘ ∠ A D C + ∠ A B C = 180 ∘ ⇒ 140 ∘ + ∠ A B C = 180 ∘ ⇒ ∠ A B C = 180 ∘ − 140 ∘ ∴ ∠ A B C = 40 ∘ Since ∠ A C B is an angle which lies in a semi – circle, ∠ A C B = 90 ∘ I n Δ A B C , ∠ B A C + ∠ A C B + ∠ A B C = 180 ∘ [by angle sum property of a triangle] ⇒ ∠ B A C + 90 ∘ + 40 ∘ = 180 ∘ ⇒ ∠ B A C = 180 ∘ − 130 ∘ = 50 ∘

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball o Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha. Shaalaa.com has the CBSE Mathematics Mathematics for Class 9 CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. RD Sharma textbook solutions can be a core help for self-study and provide excellent self-help guidance for students. Concepts covered in Using RD Sharma Mathematics for Class 9 solutions Circles exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in RD Sharma Solutions are essential questions that can be asked in the final exam. Maximum CBSE Mathematics for Class 9 students prefer RD Sharma Textbook Solutions to score more in exams. Get the free view of Chapter 15, Circles Mathematics for Class 9 additional questions for Mathematics Mathematics for Class 9 CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

• Engineering and Architecture • Computer Application and IT • Pharmacy • Hospitality and Tourism • Competition • School • Study Abroad • Arts, Commerce & Sciences • Management and Business Administration • Learn • Online Courses and Certifications • Medicine and Allied Sciences • Law • Animation and Design • Media, Mass Communication and Journalism • Finance & Accounts (B) 50° Solution: Given, ABCD is cyclic Quadrilateral and ADC = 140° We know that the sum the opposite angles in a cyclic quadrilateral is 180°. ADC + ABC = 180° 140° + ABC = 180° ABC = 180°– 140° ABC = 40° Since, ACB is an angle in semi circle ACB = 90° In ABC BAC + ACB + ABC = 180° (angle sum property of a triangle) BAC + 90° + 40° = 180° BAC = 180°– 130° = 50° Therefore option (B) is correct.

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to: a. 80º b. 50º c. 40º d. 30º Solution: It is given that ABCD is a ∠ADC = 140º Sum of the opposite angles in a cyclic quadrilateral is 180º ∠ADC + ∠ABC = 180º Substituting the values 140 + ∠ABC = 180º ∠ABC = 180º - 140º ∠ABC = 40º ∠ACB is an angle in a ∠ACB = 90º In triangle ABC Using the ∠BAC + ∠ACB + ∠ABC = 180º Substituting the values ∠BAC + 90º + 40º = 180º By further calculation ∠BAC + 130º = 180º ∠BAC = 180 - 130 ∠BAC = 50º Therefore, ∠BAC is equal to 50º. ✦ Try This: ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 120º, then ∠BAC is equal to ☛ Also Check: NCERT Exemplar Class 9 Maths Exercise 10.1 Problem 8 ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to: a. 80º, b. 50º, c. 40º, d. 30º Summary: ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to 50º ☛ Related Questions: • • •

Given: ∠DBC = 29° Concept used: Angle in a semicircle is a right angle. We know that sum of all the sides of triangles is 180° The opposite angles ina cyclic quadrilateral is 180° Calculation: According to the question ∠DBC = 29° ∠BDC = 90° [Angle in a semicircle is 90°] We know that sum of all the sides of triangles is 180° ⇒∠DBC +∠BDC +∠BCD = 180° ⇒ 29° + 90° +∠BCD = 180° ⇒ 119° +∠BCD = 180° ⇒∠BCD = (180° –119°) ⇒∠BCD = 61° Now, We know that, the opposite angles in a cyclic quadrilateral is 180° So, ⇒∠BAD +∠BCD = 180° ⇒∠BAD + 61° = 180° ⇒∠BAD = (180° –61°) ⇒∠BAD = 119° ∴ The required value of∠BAD is 119°