In the figure, it is given that

(i) AB ⊥ BF and EF⊥BF
(ii) AC = BC
(iii) KD is perpendicular to BC and DE.

Find the measure of x.
(A) 75°             (B) 30°             (C) 60°            (D) 45°

Correct option is C)

Since, KD⊥BC and DE.

∠CKD=∠EDK=90°
∠ZBC+∠BCZ+∠CZB=180°
∠BCZ=180°−90°−35°=55°
Now, In △CKD
∠CKD+∠BCZ+∠KDC=180°
So, ∠KDC=180°−90°−55°=35°
∠KDE=∠CDE+∠CDK
So, ∠CDE=∠KDE+∠CDK=90°−35°=55°
In △FZD,
∠FZD+∠ZFD+∠FDZ=180°
So, ∠FDZ=180°−90°−25°=65°
Now, ∠x+∠FDZ+∠EDC=180°
So, ∠x=180°−65°−55°=60°