Question

please e me answer it's to hard​

Answer 1

Answer:

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Answer 2

angle AOB + angle BOC = 180°. {linear pair}

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-40

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35now, to find angle AOB and angle BOC substitute the 'x' value in x+25° and 3x+15°.

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35now, to find angle AOB and angle BOC substitute the 'x' value in x+25° and 3x+15°.angle AOB = x+25°

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35now, to find angle AOB and angle BOC substitute the 'x' value in x+25° and 3x+15°.angle AOB = x+25°35+25 = 60°

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35now, to find angle AOB and angle BOC substitute the 'x' value in x+25° and 3x+15°.angle AOB = x+25°35+25 = 60°angle BOC = 3x+ 15

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35now, to find angle AOB and angle BOC substitute the 'x' value in x+25° and 3x+15°.angle AOB = x+25°35+25 = 60°angle BOC = 3x+ 153× 35 + 15

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35now, to find angle AOB and angle BOC substitute the 'x' value in x+25° and 3x+15°.angle AOB = x+25°35+25 = 60°angle BOC = 3x+ 153× 35 + 15105+15 = 120°

angle AOB + angle BOC = 180°. {linear pair}(x+25)°+(3x+15)°= 180°x+25+3x+15=180°4x+40= 180°4x = 180-404x = 140x= 140/4x= 35now, to find angle AOB and angle BOC substitute the 'x' value in x+25° and 3x+15°.angle AOB = x+25°35+25 = 60°angle BOC = 3x+ 153× 35 + 15105+15 = 120°the angles :- AOB= 60°

:- AOB= 60°

:- AOB= 60° BOC= 120°