Question

to the figure, if AB divided DAC in the ratio 1:3, determine the value of....


Class 9th chapter - line and angel.......... ​

Answer 1

Answer:

Let ∠BAD = y, ∠BAC = 3y

∠BDA = ∠BAD = y (As AB = DB)

Now, ∠BAD + ∠BAC + 108° = 180° [Linear Pair]

y + 3y + 108° = 180°

4y = 72° or

y = 18°

Now, In ΔADC

∠ADC + ∠ACD = 108° [Exterior Angle Property]

x + 18° = 180°

x = 90°

I hope it helps ☺️