Question

BAD, BCE,ACF are straight lines. it is given that BA=BC,AD=AF,EB=ED.if angle BAC=z°,find the value z ​

Answer 1

Given, BAD, BCE, ACF and DEF are straight lines and BA = BC, AD = AF, EB = ED. angle BAC = z°

Let angle ABC = y° and BED = x°

In ∆ABC,

angle BAC = angle BCA = z° (given)

By angle sum property, we get

angle (BAC + BCA + ABC) = 180°

or, 2z° + y° = 180° ___ eq.(i)

In ∆BED,

EB = ED (given)

angle EBD = angle EDB = y°

angle FAD = 180° - z° ___ eq.(ii)

(Since BAD is a straight line)

In ∆ADF,

angle ADF = y° = angle DFA (given)

By angle sum property, we get

angle(ADF + DFA + FAD) = 180°

or, 2y° + 180° - z° = 180°

or, 2y° - z° = 0° ___ eq.(iii)

From eq.(i) and eq.(iii), we get

y° = 36°

In ∆BED,

angle BED = 180° - angle EBD - angle EDB

or, x = 180° - 2y°

or, x = 180° - 2(36°)

or, x = 108°

Answer:

Therefore, the value of x is 108°.