In triangle ABC,angle B = 45°,angle C= 55°, and the bisector of angle A meets BC at point 'D' .Find angle ADB and angle ADC
Answers
Answered by
2
Step-by-step explanation:
Correct options are A) and B)
Given ∠B=45,∠C=55
So, ∠A=180°−45°−55°
=80°
From △ABD,40°+45°+∠ADB=180°
⟹∠ADB=95°
From△ADC,40°+55°+∠ADC=180°
⟹∠ADC=85°.
Answered by
1
answer:
Angle A= 180°-(55+45)°
=80°
Now,
As AD is bisector
Angle BAD=1/2 Angle A = 40°
In triangle BAD,
Angle ADB
= 180°-(Angle BAD +Angle DBA)
=180°-(40+45)°
=95°
Similarly,
In triangle ACD,
Angle ADC
= 180°-(40+55)°
=85°
answer is 85°
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