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**In the following figure, ABCD is a rhombus and DCFE is a square.**If ∠ABC =56°, find:

(i) ∠DAE

(ii) ∠FEA

(iii) ∠EAC

(iv) ∠AEC

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#### Solution

ABCD is a rhombus.

⇒ AD = CD and ∠ADC = ∠ABC = 56°

DCFE is a square.

⇒ ED = CD and ∠FED = ∠EDC = ∠DCF = ∠CFE = 90°

⇒ AD = CD = ED

In ΔADE,

AD = ED

⇒ ∠DAE = ∠AED ...(i)

∠DAE + ∠AED + ∠ADE = 180°

⇒ 2∠DAE + 146° = 180° ....( Since ∠ADE = ∠EDC + ∠ADC = 90° + 56° = 146° )

⇒ 2∠DAE = 34°

⇒ ∠DAE = 17°

⇒ ∠DEA = 17° ....(ii)

In ABCD,

∠ABC + ∠BCD + ∠ADC + ∠DAB = 360°

⇒ 56° + 56° + 2 ∠DAB = 360° ....( ∵ Opposite angles of a rhombus are equal.)

⇒ 2∠DAB = 248°

⇒ ∠DAB = 124°

We know that diagonals of a rhombus, bisect its angles.

⇒ ∠DAC = `(124°)/2` = 62°

⇒ ∠EAC = ∠DAC - ∠DAE = 62° - 17° = 45°

Now,

∠FEA = ∠FED - ∠DEA

= 90° - 17° ....( From(ii) and each angle of a square is 90° )

= 73°

We know that diagonals of a square bisect its angles.

⇒ ∠CED = `(90°)/2` = 45°

So,

∠AEC = ∠CED - ∠DEA

= 45° - 17°

= 28°

Hence, ∠DAE = 17°, ∠FEA = 73°, ∠EAC = 45° and ∠AEC = 28°.

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