Question

The angles of a quadrilateral are in AP, and the greatest angle is double the least. Express the least angle in radians. [FULL SOLUTION REQUIRED][CONTENT QUALITY IS MUST]

Answer 1

Hi there!

The angles of a quadrilateral are in AP, and the greatest angle is double the least.So,

Let a be the least angle and common difference let it be d.

Angle will be a, a+d , a+2d and a +3d

(Given that the angles of a quadrilateral are in A.P)

→ a+ a+d+ a+2d+ a+3d = 360 [ •°• Property of quadrilateral ]

→ 4a + 6d = 360

Taking two as common,

2a+3d= 180-----(1)

Now, according to question, the greatest angle is double the least.

Greater one is = a+3d

Smallest one is = a

So, forming equation

a+3d (greatest angle) = 2 ×a (smallest angle) 

→a+ 3d = 2a-------(2)

From equation (2),

a= 3d

Put a in (1) equation.

2a + a = 180° 

3a = 180°

a= 60°

Now solving equation 3d = a by putting a= 60

d = 20

So, the four angles are 60°, 80°, 100°, 120°

Least angle in radian i.e 60°

60° × π/180°

=π/3

Thankyou :)

Answer 2

Answer:-

least angle = π/3

Given :-

The angles of quadrilateral are in AP.

Greatest angle is double the least.

To express :-

The least angle is radians.

Solution:-

Let the angles in AP be a+2d , a - 2d , a + d and a - d.

A/Q

→$\mathsf{a-2d + a-d + a+d + a+2d = 360^{\circ}}$

→$\mathsf{4a = 360^{\circ}}$

→$\mathsf{a = \dfrac{360}{4}}$

→$\mathsf{ a = 90^{\circ}}$

Now,

→$\mathsf{a+2d = 2 (a-2d)}$

→$\mathsf{ a + 2d = 2a -4d}$

→$\mathsf{a-2a = -4d -2d}$

→$\mathsf{-a = -6d}$

→$\mathsf{-6d = -90}$

→$\mathsf{ d = \dfrac{-90}{-6}}$

→$\mathsf{d = 15}$

Hence,

The four angle of a quadrilateral in A. P will be :-

→a - 2d

= 90 - 2 × 15

=90 - 30

= 60°

→a - d

= 90 - 15

= 75°

→a + d

= 90 + 15

= 105°

→a + 2d

= 90 + 2 × 15

= 90 + 30

= 120°

Least angle in radian is = 60°

→$\mathsf{\left(60 \times \dfrac{\pi}{180}\right)}$

→$\dfrac{\pi}{3}$

hence,

The least angle is radian is π/3.