Question

3 sum geometry

plz help need tommorow

plz

Answer 1

Answer:

x:140 degrees

y:110 degrees

z:100 degrees

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Answer 2

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a) Given

• P(AB) // Q(CD)
• Line m(GH) is transversal
• Angle HEB = 40°

To Find:

• Value of x

Solution:

$As\:P//Q\:and\:m\:is\:transversal \\ \implies Angle\: HEB\:=\:Angle \:EFD\:=\:40^{\circ} -----(Corresponding\:Angles) \\ \implies Angle\: EFC \:+ \:Angle\:EFD\:=\:180^{\circ} -----(CD\:is\:a\:straight\:line)\\ \implies x\:+\:40^{\circ}\:=\:180^{\circ} \\ \implies\bold{ x\:=\:140^{\circ} }$

b) Given:

• Angle ACD = 40°
• Angle ECB = 30°
• AB is a straight line

To Find:

• Value of y

Solution:

$As\:AB\:is\:a\:straight\:line \\ \implies Angle \:ACD \:+\: Angle \:DCE + Angle \:ECB\: = \: 180^{\circ} \\ \implies 40^{\circ} + y + 30^{\circ}\:=\: 180^{\circ} \\ \implies y\: =\: 180^{\circ}\:-\: 70^{\circ} \\ \implies \bold{ y \: =\: 110^{\circ} }$

c) Given:

• Angle SPT = 60°
• Angle QTR = 20°

To Find:

• Value of z

Solution:

$\implies Angle \:STP\:=\: Angle \:RTQ \:=\:20^{\circ} -----(Vertically\: opp.\: angles) \\ In\:\Delta\:PTS\\ \implies Angle \:STP\:+ \:Angle \:SPT\: +\:Angle \:PST \:=\:180^{\circ} ------(Angle \:sum\: of\: Triangle) \\ \implies 20^{\circ}\:+\: 60^{\circ} \:+\:z\:=\:180^{\circ} \\ \implies z\:=\:180^{\circ}\:-\:80^{\circ} \\ \implies \bold{ z\:=\:100^{\circ}}$

Hope it helps!!!