Question

please give me solution step by step it is urgent​

Answer 1

Answer:

1) <HIG

(15°+90°) - 180° = <HIG (angle sum property)

105-180° = <HIG

<HIG = 75°

2) <JIF

FG is a straight line

<HIG + <JIH = <JIF

75° + 50° = <JIF

<JIF = 125°

3) <IJF

JFI is a triangle

(<JFI + <JIF) - 180° = <IJF (angle sum property)

(90°  + 125°) - 180°  = <IJF

215° - 180° = <IJF

<IJF = 35°

4) <HIF

<JIH + <JIF = < HIF

50° + 125°  = <HIF

<HIF = 175°

5) <JHI

(<HJI + <JIH) - 180° (angle sum property)

first we have to find <HJI => [<EFH + <IJF - 180°]

                                 <HJI => [60° + 35° - 180°]

                                 <HJI =>   [95 - 180°]

                                  <HJI =>  85°

<JHI = (<HJI + <JIH) - 180°

<JHI = (85° + 50°) -180°

<JHI = 135 - 180°

<JHI = 45°

6) <JHE

(<HEJ + <EJH) - 180° (angle sum property)

(90° + 60°) - 180° = <JHE

150° - 180° = <JHE

<JHE = 30°