i need answers for the problems in the image above,as soon as possible.
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7. angle AOB =110°( since OP bisects the angle AOP)
8.angle NLO=angle NMO = 25°
angle NLO+angle NMO =angle LKM= 50°
OK bisects the angle LKM
Therefore Angle LKN =25°
9.In triangle ADB
angle A =50°
angle D = 60°( since angle D = angleC)
Therefore Angle ABD = 180 - angle A + angle D - ( sum of the angles in a triangle)
=180- 50+60
= 70°
plz mark as branliest answer.
8.angle NLO=angle NMO = 25°
angle NLO+angle NMO =angle LKM= 50°
OK bisects the angle LKM
Therefore Angle LKN =25°
9.In triangle ADB
angle A =50°
angle D = 60°( since angle D = angleC)
Therefore Angle ABD = 180 - angle A + angle D - ( sum of the angles in a triangle)
=180- 50+60
= 70°
plz mark as branliest answer.
MSMS:
Bt its nt mentioned tat OP bisects angle AOP.
It is nothing but a line drawn centre of a circle to a chord bisects the chord and the angle.
Actually fr ths theorem to be applicable it must have to be perpendicular... Ny line from centre won't bisect the chord.
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Here's the solution to your question......
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