RRB Junior Engineer Stage 1 Model Papers 1

Question : 5 of 100

Marks: +1, -0

In a triangle ABC, AB = AC, BA is produced to D in such a manner that AC = AD. The circular measure of

∠BCD

Solution:

AB = AC

∴ ∠ABC = ∠ACB

... (1)

[opposite angle of equal sides are equal]

AC = AD

∴ ∠ACD = ∠ADC

... (2)
In a triangle,

∠ABC + ∠ADC + ∠DCB = 180°

∠ABC + ∠ADC + ∠ACB + ∠ACD = 180°

2 ∠ACB + 2 ∠AOC = 180°

[From eqn. (1) & (2) ]

∴ ∠BCD = 90° \or π/2

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