Question

please answer my questions

Answer 1

$\color{violet} {\underline{\bold{Hey \: here \: is \: your \: answer \: => }}}$

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$\color{blue} {\boxed{\underline{\huge{\mathbb{\star{GIVEN \: }\star}}}}}$

$<b >EF is a line.$

$\color{blue} {\boxed{\underline{\huge{\mathbb{\star{SolutiOn \: }\star}}}}}$

On line EF

=> <EBD + <DBF = 180° ( linear pair )
=> <EBD + <DBC + <CBF = 180° ( <DBF = <DBC + <CBF )
=> x + 3° + x + 20° + x + 7° = 180°
=> 3x + 30° = 180°
=> 3x = 180° - 30°
=> 3x = 150°
=> x = 150° / 3
=> x = 50°

So x = 50°

and all angles of figure are

(I) x + 3° = 50° + 3° = 53°
(ii) x + 20° = 50° + 20° = 70°
(iii) x + 7° = 50° + 7° = 57°

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Hope this helps you

### BE BRAINLY ###

Answer 2

Hi ,

From. the figure ,

<EBD + <DBC + <CBF = 180°

[ Straight angle ]

( x + 3 )° + ( x + 20 )°+ ( x + 7 )° = 180°

x + 3 + x + 20 + x + 7 = 180°

3x + 30° = 180°

3x = 180° - 30°

3x = 150°

x = 150/3

x = 50

Therefore ,

<EBD = ( x + 3 )° = ( 50 + 3 )° = 53°

<DBC = ( x + 20 )° = ( 50 + 20 )° = 70°

<CBF = ( x + 7 )° = ( 50 + 7 )° = 57°

I hope this helps you.

: )