Question

The angle of a quadrilateral are in A.P and the greatest angle is double the least find the angle of the quadrilateral in radian​

Answer 1

Step-by-step explanation:

Let a-3d, a-d, a+d & a+3d be the angles of quadrilateral in AP.

$\because$ Sum of all the angles of a quadrilateral is 360°

$\therefore \: a - 3d + a - d + a + d + a + 3d = 360 \degree \\ \therefore \: 4a = 360 \degree \\ \therefore \: a = \frac{360 \degree}{4} \\ \therefore \: a = 90 \degree \\ Here \\ Greatest \:angle = a + 3d\\ Least \:angle = a - 3d \\ According \: to \: the \: given \: condition: \\ a + 3d = 2(a - 3d) \\ a + 3d = 2a - 6d \\ 3d + 6d= 2a - a \\</p><p>9d= a \\</p><p>9d= 90 \degree \\</p><p>\implies d= 10 \degree\\</p><p>\implies a - 3d=90 \degree - 30 \degree=60 \degree\\</p><p>60 \degree= 60\times \frac{\pi^c}{180} =\frac{\pi^c}{3} \\</p><p></p><p>\implies a - d=90\degree - 10 \degree=80\degree\\</p><p>80 \degree= 80\times \frac{\pi^c}{180} =\frac{4\pi^c}{9}\\ </p><p></p><p>\implies a + d=90\degree + 10 \degree=100\degree\\</p><p>100 \degree= 100\times \frac{\pi^c}{180} =\frac{5\pi^c}{9} \\</p><p></p><p>\implies a +3d=90 \degree +30 \degree=120 \degree\\</p><p>120 \degree= 120\times \frac{\pi^c}{180} =\frac{2\pi^c}{3}$

Thus the measures of the angles of quadrilateral in radian are:

$\frac{\pi^c}{3}, \frac{4\pi^c}{9}, \frac{5\pi^c}{9}\: and \: \frac{2\pi^c}{3}$

Answer 2

Answer:Let a-3d, a-d, a+d & a+3d be the angles of quadrilateral in AP.

Sum of all the angles of a quadrilateral is 360°

Thus the measures of the angles of quadrilateral Let a-3d, a-d, a+d & a+3d be the angles of quadrilateral in AP.

Sum of all the angles of a quadrilateral is 360°

Thus the measures of the angles of quadrilateral in radian are: