Calculate x :

(i)

(ii)

(i) In △ ABC,

⇒ BC = AC (Given)

∴ ∠BAC = ∠CBA = 37° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠BAC + ∠CBA + ∠ACB = 180°

⇒ 37° + 37° + ∠ACB = 180°

⇒ ∠ACB + 74° = 180°

⇒ ∠ACB = 180° - 74° = 106°.

From figure,

Since, BCD is a straight line,

∴ ∠ACB + ∠ACD = 180°

⇒ 106° + ∠ACD = 180°

⇒ ∠ACD = 180° - 106° = 74°.

In △ ACD,

⇒ CD = AD (Given)

∴ ∠CAD = ∠ACD = 74° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠CAD + ∠ACD + ∠ADC = 180°

⇒ 74° + 74° + x = 180°

⇒ 148° + x = 180°

⇒ x = 180° - 148° = 32°.

Hence, x = 32°.

(ii) In △ ABC,

⇒ BC = AC (Given)

∴ ∠BAC = ∠CBA = 50° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠BAC + ∠CBA + ∠ACB = 180°

⇒ 50° + 50° + ∠ACB = 180°

⇒ ∠ACB + 100° = 180°

⇒ ∠ACB = 180° - 100° = 80°.

From figure,

Since, BCD is a straight line,

∴ ∠ACB + ∠ACD = 180°

⇒ 80° + ∠ACD = 180°

⇒ ∠ACD = 180° - 80° = 100°.

In △ ACD,

⇒ CD = AC (Given)

∴ ∠CAD = ∠ADC = x (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠CAD + ∠ACD + ∠ADC = 180°

⇒ x + x + 100° = 180°

⇒ 2x + 100° = 180°

⇒ 2x = 180° - 100°

⇒ 2x = 80°

⇒ x = $\dfrac{80°}{2}$ = 40°.

Hence, x = 40°.