Question

The angles of the quadri lateral are in an Ap. The greatest angle is
double of the smallest angle find
the four angles of quadrilateral​

Answer 1

Answer:

60°, 80°, 100°, 120° .

Step-by-step explanation:

Let the angles are a - 3d, a - d, a + d and a + 3d, where a(>0) is the first term and 2d(>0) is the common difference.

According to question :

= > greatest angle = 2 * smallest angle

= > a + 3d = 2( a - 3d )

= > a + 3d = 2a - 6d

= > 3d + 6d = 2a - a

= > 9d = a ... (1)

Sum of all angles at vertices, in quadrilateral, is 360°.

= > ( a - 3d ) + a - d + a + d + ( a + 3d ) = 180°

= > a - d + a - d + a + d + a + 3d = 180°

= > 4a = 360°

= > a = 90°.

Therefore,

= > 9d = a = 90°

= > 9d = 90°

= > d = 10°.

Hence, angles are -

• a - 3d = 90° - 3(10)° = 60°

• a - d = 90° - 10° = 80°

• a + d = 90° + 10° = 100°

• a + 3d = 90 + 30 = 120

Answer 2

Given ,

The angles of the quadri lateral are in an AP

The greatest angle is double of the smallest angle

Let , The angles of quadrilateral be :

" a - 3d " ," a - d " , " a + d " and " a + 3d "

We know that , the sum of all angles of quadrilateral is 360

a - 3d + a - d + a + d + a + 3d = 360

4a = 360

a = 360/4

a = 90

Since , the greatest angle is double of the smallest angle

Thus ,

a + 3d = 2(a - 3d)

a + 3d = 2a - 6d

a = 9d

90 = 9d

d = 90/9

d = 10

$\therefore$ The angles of quadrilateral are 60 , 80 , 100 and 120