MissArixseidn
Answered

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Euclid Street branches out into Pythagoras Street and Hipparchus Street such that Pythagoras Street is perpendicular to Hipparchus Street. On the right side, Pythagoras Street makes an angle of 150° with Euclid Street. Solve for the values of x, y,and z a shown in the figure.

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Euclid Street Branches Out Into Pythagoras Street And Hipparchus Street Such That Pythagoras Street Is Perpendicular To Hipparchus Street On The Right Side Pyth class=

Sagot :

Answer:

x=180

180-150=30

y=60

X=120

120-60=60

60-30=30

Step-by-step explanation:

KyrieAeliaidn

Euclid Street branches out into Pythagoras Street and Hipparchus Street such that Pythagoras Street is perpendicular to Hipparchus Street. On the right side, Pythagoras Street makes an angle of 150° with Euclid Street. Solve for the values of x, y,and z a shown in the figure.

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Euclid Street is a straight angle so it's 180°.

  • [tex] \large \bold{Solve \: \: for \: \: x. } \: \boxed{ \begin{array}{} \tt x = 180\degree - 150 \degree \\ \tt \green {x = 30 \degree} \end{array}}[/tex]

  • [tex] \bold{Solve \: \: for \: \: y. } \: \boxed{ \begin{array}{} \tt y = 150 \degree - 30 \degree \\ \green{ \tt x = 60 \degree} \end{array}}[/tex]
  • [tex] \large \bold{Solve \: \: for \: \: z. } \: \boxed{ \begin{array}{} \tt z = 180 \degree - 60 \degree \\ \green{ \tt z = 120 \degree}\end{array}}[/tex]

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