Question

calculate the size of each lettered angle in the following figures. ( Please ignore the pencil working that my teacher made me write) ​

Answer 1

Answer:

x=93° (by linear pair)

z=33°(by linear pair and angle sum property of a triangle)

y=33°(by linear pair and then angle sum property of a triangle)

Hope this helps..

Answer 2

Answer:

I have given Naming in Attachment

Line ACB and Line ACE ( X ) formed a linear pair

So,

$\implies acb + ace = {180}^{0}$

$\implies {87}^{0} + x = {180}^{0}$

$\implies x = {180}^{0} - {87}^{0}$

$\implies x = {93}^{0}$

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Line ABD and Line ABC formed a Linear Pair

So,

$\implies abd + abc = {180}^{0}$

$\implies abd + {75}^{0} = {180}^{0}$

$\implies abd = {180}^{0} - {75}^{0}$

$\implies abd = {105}^{0}$

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Line AEF and Line AEC formed a Linear Pair

So,

$\implies aef + aec = {180}^{0}$

$\implies {126}^{0} + aec = {180}^{0}$

$\implies aec = {180}^{0} - {126}^{0}$

$\implies aec = {54}^{0}$

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By Angle Sum Property ,

$\implies aec + ace + cae(y) = {180}^{0}$

$\implies {54}^{0} + {93}^{0} + y = {180}^{0}$

$\implies {147}^{0} + y = {180}^{0}$

$\implies y = {180}^{0} - {147}^{0}$

$\implies y = {33}^{0}$

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By Angle Sum Property ,

$\implies bad(z) + abd + adb = {180}^{0}$

$\implies z + {105}^{0} + {42}^{0} = {180}^{0}$

$\implies z + {147}^{0} = {180}^{0}$

$\implies z = {180}^{0} - {147}^{0}$

$\implies z = {33}^{0}$

Hence ,

Value of X = 93°

Value of Y = 33°

Value of Z = 33°

HØPÈ ÏT HÊLPS ♠♦♠