Question

Tha angles of a quadrilateral are in A.P., whose common difference is ${10}^o$. Find the angles.

Answer 1

Given,

the angles of a quadrilateral are in A.P, whose common difference $d={10}^0$

let the first angle of the quadrilateral be $a......eq\left(1\right)$

hence second angle must be $a+{10}^0\dots .eq\left(2\right)$

similarly,

third angle is $a+{10}^0+{10}^0=a+{20}^0\dots .eq\left(3\right)$

fourth angle is $a+{30}^0\dots ...eq\left(4\right)$

now, we know that sum of all the angles of a quadrilateral must be ${360}^0$

hence

$a+a+{10}^0+a+{20}^0+a+{30}^0={360}^0$

$⟹4a+{60}^0={360}^0$

$⟹4a={360}^0-{60}^0$

$⟹4a={300}^0$

$a={75}^0$ first angle.

now put value of a in eq(2). eq(3) and eq(4 we get)

second angle ${75}^0+{10}^0={85}^0$

third angle ${75}^0+{20}^0={95}^0$

fourth angle ${75}^0+{30}^0={105}^0$

hence the A.P for these angles is

${75}^0,{85}^0,{95}^0,{105}^0$