Question

In a triangle ABC, angle B=80 degrees, angle C=20 degrees. A line is drawn to any point D on BC.Such that AB=CD, Find the measure of angle BDA. The correct answerer will get 60 points and I'll mark him brainliest.​

Answer 1

Answer:

20° < ∠BDA <  100°

Step-by-step explanation:

In a triangle ABC, ∠B=80°, ∠C=20°

∠A + ∠B + ∠C = 180°

=> ∠A + 80° + 20° = 180°

=> ∠A = 80°

A line is drawn to any point D on BC

=> ∠BDA  = ∠DAC +  ∠C

=> ∠BDA  = ∠DAC  + 20°

∠DAC > 0 °

=> ∠BDA  > 20°

∠BDA  = ∠DAC  + 20°

∠DAC = ∠A -  ∠BAD

=> ∠DAC  = 80° - ∠BAD

∠BDA  =  80° - ∠BAD + 20°

=>  ∠BDA  =   100° - ∠BAD

∠BAD > 0°

=> ∠BDA <  100°

20° < ∠BDA <  100°

finally

80+20+a=180

a = 180-100=80

ab=dc

80×80=20×d

d = (80×80)/20

d = 320

Answer 2

Answer: Measure of ∠BDA is 60°

Solution:

• ∠B = 80° & ∠C = 20°

• ∠A + ∠B + ∠C = 180°  (∵  Angle sum property of triangle)

• ∠A + 80 +20 = 180°

• ∠A = 80°

• Now, lets say that AD is such that it is angle bisector of ∠A

• ∠BAD = 80/2=40°

• ∠BAD+ ∠B + ∠BDA = 180°( Angle sum property of triangle)

• 40+80 + ∠BDA = 180°

• ∠BDA = 60°

Hope It Helps!!

Plzzzzz Mark As Brainliest!!