1.

To the figure, if AB divided DAC in the ratio 1:3, determine the value of....Class 9th chapter - line and angel.......... ​

Answer»

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 ☺️


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