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In a cyclic quadrilateral ABCD, ∠A = (x + 7)°, ∠B = (y + 8)°, ∠C = (3y + 23)° and ∠D = (4x + 12)°. Find all four angles of the cyclic quadrilateral.
In a cyclic quadrilateral, opposite angles are supplementary. Thus, ∠A + ∠C = (x + 7) + (3y + 23) = 180°, which simplifies to x + 3y = 150. Also, ∠B + ∠D = (y + 8) + (4x + 12) = 180°, giving 4x + y = 160. Multiplying the first equation by 4 and subtracting the second eliminates x, yielding 11y = 440Read more
In a cyclic quadrilateral, opposite angles are supplementary. Thus, ∠A + ∠C = (x + 7) + (3y + 23) = 180°, which simplifies to x + 3y = 150. Also, ∠B + ∠D = (y + 8) + (4x + 12) = 180°, giving 4x + y = 160. Multiplying the first equation by 4 and subtracting the second eliminates x, yielding 11y = 440, so y = 40 and x = 30. The angles are 37°, 48°, 143° and 132°.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessThe sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.
Let the tens digit be x and the units digit be y, making the original number 10x + y. The sum of digits gives x + y = 15. Reversing the digits gives 10y + x. The difference condition (10y + x) − (10x + y) = 9 reduces to 9y − 9x = 9 or y − x = 1. Adding x + y = 15 and y − x = 1 yields 2y = 16, givingRead more
Let the tens digit be x and the units digit be y, making the original number 10x + y. The sum of digits gives x + y = 15. Reversing the digits gives 10y + x. The difference condition (10y + x) − (10x + y) = 9 reduces to 9y − 9x = 9 or y − x = 1. Adding x + y = 15 and y − x = 1 yields 2y = 16, giving y = 8 and x = 7. The number is 78.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessAt a certain time, Jacob notices that his digital watch reads a minutes after two o’clock. Fifteen minutes later, it reads b minutes after three o’clock. He noticed that a is six times greater than b. What time was it when he looked at his watch for the second time?
The time elapsed from a minutes past two to b minutes past three is 15 minutes. Expressing both in minutes after two o'clock gives the relation (60 + b) − a = 15, which simplifies to a − b = 45. Given that a = 6b, substituting gives 6b − b = 45, leading to 5b = 45 or b = 9. Therefore, Jacob's watchRead more
The time elapsed from a minutes past two to b minutes past three is 15 minutes. Expressing both in minutes after two o’clock gives the relation (60 + b) − a = 15, which simplifies to a − b = 45. Given that a = 6b, substituting gives 6b − b = 45, leading to 5b = 45 or b = 9. Therefore, Jacob’s watch read 3:09 at the second viewing.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessRobot 1 starts from (3, 0) and traces a path by repeatedly moving 5 units to the right and then 5 units upward. Robot 2 starts from (7, 0) and repeatedly moves 5 units to the right and then 3 units downwards. Will the paths traced by these two robots intersect and if so, where?
The paths will not intersect. Robot 1 starts at (3, 0) and moves upward into the first quadrant, having y = x − 3 > 0 for all x > 3. Robot 2 starts further right at (7, 0) and moves downward, having y = −(3/5)(x − 7) ≤ 0 for all x ≥ 7. Because Robot 1 is always strictly positive where Robot 2Read more
The paths will not intersect. Robot 1 starts at (3, 0) and moves upward into the first quadrant, having y = x − 3 > 0 for all x > 3. Robot 2 starts further right at (7, 0) and moves downward, having y = −(3/5)(x − 7) ≤ 0 for all x ≥ 7. Because Robot 1 is always strictly positive where Robot 2 travels, the two traced paths never meet.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See lessRobot 1 starts from the origin and traces a path by repeatedly moving 3 units to the right and then 4 units upward. Robot 2 starts from (10, 0) and repeatedly moves 1 unit to the right and then 2 units upward. Will the paths traced by these two robots intersect and if so, where?
Robot 1 moves with slope 4/3 from (0, 0), following the linear path y = (4/3)x. Robot 2 moves with slope 2 starting from (10, 0), following y = 2x − 20. Solving simultaneously by substitution gives 2x − 20 = (4/3)x, leading to (2/3)x = 20, so x = 30. Substituting gives y = 40. Both robots reach (30,Read more
Robot 1 moves with slope 4/3 from (0, 0), following the linear path y = (4/3)x. Robot 2 moves with slope 2 starting from (10, 0), following y = 2x − 20. Solving simultaneously by substitution gives 2x − 20 = (4/3)x, leading to (2/3)x = 20, so x = 30. Substituting gives y = 40. Both robots reach (30, 40) in discrete integer steps, so their paths intersect at (30, 40).
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 13 Two Variables, One Line Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-13/
See less