Question

the angle of a quadrilateral are in A.P with d=10°. find angles​

Answer 1

Step-by-step explanation:

angles be :a-d,a,a+d,a+2d

then in quad.

a-d+a+a+d+a+2d =360°

4a+2d=360°

4a+2*10=360. (given d=10°)

4a+20=360

4a=340

a =340/4

a= 85°

therefore angles : 75°,85°,95°,105°

Answer 2

The angles are $\boxed{75^\circ, 85^\circ, 95^\circ, 105^\circ}$

Step-by-step explanation:

Let the smallest angle of the quadrilateral be a

Then the angles of the quadrilateral will be , given that the angles are in AP with common difference d

$a, a+d, a+2d, a+3d$

We know that sum of all the angles of a quadrilateral = 360°

Thus,

$a+(a+d)+(a+2d)+(a+3d)=360^\circ$

$\implies 4a+6d=360^\circ$

$\implies 2a+3d=180^\circ$

But given that $d=10^\circ$

Therefore,

$2a+3\times 10^\circ=180^\circ$

$\implies 2a+30^\circ=180^\circ$

$\impllies 2a=150^\circ$

$\implies a=75^\circ$

Therefore, the angles are

$75^\circ, 75^\circ+10^\circ, 75^\circ+2\times 10^\circ, 75^\circ+3\times 10^\circ$

or, $75^\circ, 85^\circ, 95^\circ, 105^\circ$

Hope this answer is helpful.

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