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

(i) ∠DAG

(ii) ∠FEG

(iii) ∠GAC

(iv) ∠AGC.

(i) We know that,

Each angle of a square = 90°.

As ABCD is a rhombus so, AB = BC = DC = AD …….(i)

Also, CD = ED = FC = EF (As CDEF is a square) … (ii)

From (i) and (ii), we have

AB = BC = DC = AD = EF = FC = ED … (iii)

∠ABC = 56° [Given]

⇒ ∠ADC = ∠ABC = 56° [Opposite angle in rhombus are equal]

From figure,

⇒ ∠EDA = ∠EDC + ∠ADC = 90° + 56° = 146°

In ∆ADE,

⇒ DE = AD [From (iii)]

⇒ ∠DEA = ∠DAE [Equal sides have equal opposite angles]

From figure,

⇒ ∠DEA = ∠DAE = ∠DAG

⇒ ∠DAE + ∠DEA + ∠EDA = 180°

⇒ ∠DAG + ∠DAG + ∠EDA = 180°

⇒ 2∠DAG + 146° = 180°

⇒ 2∠DAG = 180° - 146° = 34°

⇒ ∠DAG = $\dfrac{34°}{2} = 17°$

⇒ ∠DAG = 17°.

Hence, ∠DAG = 17°.

(ii) Also,

⇒ ∠DEG = 17°

⇒ ∠FEG = ∠E – ∠DEG

= 90° – 17°

= 73°

Hence, ∠FEG = 73°.

(iii) In rhombus ABCD,

⇒ ∠A + ∠B = 180° (As AD || BC, the sum of co-interior angles = 180°.)

⇒ ∠A = 180° - ∠B = 180° - 56° = 124°

⇒ ∠DAC = $\dfrac{124°}{2} = 62°$ [∵ Diagonal AC bisects ∠A]

⇒ ∠DAC = 62°

⇒ ∠GAC = ∠DAC – ∠DAG

= 62° – 17° = 45°.

Hence, ∠GAC = 45°.

(iv) In ∆EDG,

⇒ ∠D + ∠DEG + ∠DGE = 180° [Angles sum property of a triangle]

⇒ 90° + 17° + ∠DGE = 180°

⇒ ∠DGE = 180° – 107° = 73°

Thus, ∠AGC = ∠DGE [Vertically opposite angles are equal]

⇒ ∠AGC = 73°.

Hence, ∠AGC = 73°.