In △ ABD and △ FEC, ∠BAD = 60°, l(BD) = l(EC), ∠ABD = ∠FEC = 90°, and l(AB) = l(FE). Find the ratio of ∠BADover ∠FCE . [Image]

In △ ABD and △ FEC, ∠BAD = 60°, l(BD) = l(EC), ∠ABD = ∠FEC…

In $\triangle$ ABD and $\triangle$ FEC, ∠BAD = 60°, l(BD) = l(EC), ∠ABD = ∠FEC = 90°, and l(AB) = l(FE). Find the ratio of $\angle{BAD}\over \angle{FCE}$ .

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Correct answer: D

Explanation

The correct answer is option 4 i.e. 2 : 1.

In triangle ABD,

∠BAD + ∠ABD + ∠BDA = 180°

⇒ 60° + 90° + ∠BDA = 180°

⇒ ∠BDA = 180° - 150° = 30°

In triangle ABD and triangle FEC

AB = EF, ∠ABD = ∠FEC = 90°, BD = EC

△ABD and △FEC are congruent (Side angle side)

∠BDA = ∠FCE = 30°

So, $\angle{BAD}\over \angle{FCE}$ = 60/30 = 2 : 1

SSC CGL 2024 PYQ — Question 68