# Please find the value of x. Q. In the given figure, find the value of x ? lines and angles

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I am solving this .
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101

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The value of x is 109
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Dear student,
According to the figure,

In triangle ABC,
angle ABC + angle BAC + angle CAB = 180' (angle sum property)
angle ABC + 34' + 30' = 180'
angle ABC = 180' - 34' + 30'
angle ABC = 116'

Then, we can see that,
angle ABC + angle CBD = 180' (linear pair)
116' + angle CBD = 180'
angle CBD = 180' - 116'
angle CBD = 64'

In triangle EBD, we can see that,
angle EDB + angle DBE + angle BED = 180' (angle sum property)
45' + 64' + angle DBE = 180'
angle DBE = 180' - 45' - 64'
angle DBE = 71'

And now we can identify that,
angle DBE + angle CDE = 180' (linear pair)
71' + x' = 180'
x' = 180' - 71'
x' = 109'

Therefore we have found out that x' = 109'

Hope you have understood!!
Cheers!!
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angle    A  +  angle  C   =    angle  CBD     ( sum of interior opposite angle is equal opposite  exterior angle  )

= 30 +34 = 64
= angle  CBD   = 64

x  =  angle CBD + D

x  =64 +45    = 109                               (    "             "         "                 "                 "            "              "            "  )
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other wise

sum  of all  angles of an triangle = 180

a +  c +   cba    =  180

34 + 30 +cba     = 180

64 + cba     = 180

cba     =  180 -64

=  116

angle cba  + angle  cbd  =180    ( linear pair)

116 +cbd  = 180

cbd   =  180 -116

cbd  = 64

x   = angle  cbd  + d

x =   64 +  45                     ( sum of interior opposite angle is equal to opposite exterior angle   )

angle x  =  109
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109
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In triangle ABC
34 + 30 + angle CBA = 180 (angle sum property of a triangle)
64 + angle CBA = 180
angle CBA = 180 - 64
Therefore angle CBA = 116

Angle EBA + Angle EBD = 180 (Linear Pair)
116 + Angle EBD = 180
Angle EBD = 180 - 116
Therefore Angle EBD = 64

In triangle EBD
Exterior angle x = Angle EBD + Angle EDB (exterior angle is equal to the sum of it's two opposite interior                                                                               angles)
Exterior angle x = 64 + 45
Therefore, Exterior angle x = 109
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According to the question ∆ABC =A-34°,B-y,C-30° =A+B+C=34°+30°+y°=180° =64°+y°=180° y°=180°-64°=116° so:z+y=180°(Linear pair) z+116°=180° =180°-116°=64°. z=64° =∆BED=z+D+E=180°(angle sum properties of∆) =64°+45°+E=180°. 180°-109°=71° E=71° So,x= E+x=180°(Linear pair) x=180°-71°=109° x=109° Value of x=109°
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