Question

Quas. The angles of a quadrilateral are in AP with d=10. Find the angles​

Answer 1

Step-by-step explanation:

given= 10°

let us take angles as a-d , a , a+d , a+2d

now we know that sum of all quadrilateral is 360°

so we get

a-d + a + a+d + a+2d= 360°

4a+ 2(10) = 360°

a= 85°

first angle = a-d = 85 - 10= 75°

second angle a= 85°, third angle = a+2d= 85+ 2(10)= 85+20= 105°

fourth angle = a+d = 85+ 10= 95°

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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