Question

Hello I have a chemistry end of term test It's an emergencyyy heeelp !

Answer 1

Answer:

Since AOB is a straight line, we have

ZAOE + LBOE = 180°

= 3x° +72⁰º = 180°

3x° 108⁰x36⁰ ⇒ = =

We also know that

:: ZAOC + ZCOD + LBOD = 180°

straight angle)

→ æ° +90° + y = 180° +yo

⇒ 36° +90° + y = 180°

yº = 180° 126⁰ = 54°

:. ZAOC = 36°, ZBOD = 54° and 9

ZAOE 108⁰ =Since AOB is a straight line, we have

ZAOE + LBOE = 180°

= 3x° +72⁰º = 180°

3x° 108⁰x36⁰ ⇒ = =

We also know that

:: ZAOC + ZCOD + LBOD = 180°

straight angle)

→ æ° +90° + y = 180° +yo

⇒ 36° +90° + y = 180°

yº = 180° 126⁰ = 54°

:. ZAOC = 36°, ZBOD = 54° and 9

ZAOE 108⁰ =Since AOB is a straight line, we have

ZAOE + LBOE = 180°

= 3x° +72⁰º = 180°

3x° 108⁰x36⁰ ⇒ = =

We also know that

:: ZAOC + ZCOD + LBOD = 180°

straight angle)

→ æ° +90° + y = 180° +yo

⇒ 36° +90° + y = 180°

yº = 180° 126⁰ = 54°

:. ZAOC = 36°, ZBOD = 54° and 9

ZAOE 108⁰ =