Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The angles of a cyclic quadrilateral ABCD are angleA = (6x + 10)°, angleB = (5x)° angleC = (x + y)°, angleD = (3y – 10)° Find x and y, and hence the values of the four angles.

The angles of a cyclic quadrilateral ABCD are \angleA = (6x + 10)°, \angleB = (5x)° \angleC = (x + y)°, \angleD = (3y – 10)°
Find x and y, and hence the values of the four angles.

Answers (1)

Solution:
According to property of cyclic quadrilateral, sum of opposite angles = 180^{\circ}.
\angleA + \angleC = 180^{\circ}
6x + 10^{\circ} + x + y = 180^{\circ}
7x + y = 180^{\circ} –10^{\circ}
7x + y = 170^{\circ}              … (1)
Similarly \angleB + \angleD = 180^{\circ}
5x + 3y – 10^{\circ} = 180^{\circ}
5x + 3y = 180^{\circ} + 10^{\circ}
5x + 3y = 190^{\circ}           … (2)
Solving eq. (1) and eq. (2) we get
21x + 3y = 510^{\circ} {Multiply eq. (1) by 3}        … (3)
Now subtract equation (2) from (3) then we get
16x = 320^{\circ}

x= \frac{320^{\circ}}{16}
x = 20^{\circ}
Put x = 20^{\circ} in eq. (1) we get
7(20^{\circ}) + y = 170^{\circ}
y = 170^{\circ} – 140^{\circ}
y = 30^{\circ}
Hence \angleA = 6x + 10^{\circ}
= 6 × 20^{\circ} +10^{\circ}
=120^{\circ} + 10^{\circ}
= 130^{\circ}
\angleB = (5x)
 = 5 × 20^{\circ} = 100^{\circ}
\anglec= \left ( x+y \right )^{\circ}
(20^{\circ} + 30^{\circ}) = 50^{\circ}
\angleD = (3y – 10)°
= 3 × 30^{\circ} –10^{\circ}
= 90^{\circ} –10^{\circ}
= 80^{\circ}
Hence the value of four angles are:
130^{\circ}, 100^{\circ}, 50^{\circ} and 80^{\circ} respectively.

Posted by

infoexpert27

View full answer

Similar Questions

Latest Question