In each of the following figures, find the value of x :

(i)

(ii)

(iii)

(i) From figure,

∠A = 40°

In △ABC,

AB = AC

⇒ ∠C = ∠B = z (let) (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠A + ∠B + ∠C = 180°

⇒ 40° + z + z = 180°

⇒ 2z = 180° - 40°

⇒ 2z = 140°

⇒ z = $\dfrac{140°}{2}$

⇒ z = 70°

⇒ ∠C = ∠B = 70°.

From figure,

⇒ ∠C + ∠ACD = 180° (Linear pair)

⇒ 70° + x° = 180°

⇒ x° = 180° - 70°

⇒ x° = 110°

⇒ x = 110.

Hence, the value of x = 110.

(ii) From figure,

In △CAD,

⇒ AC = CD

⇒ ∠CDA = ∠CAD = 30° (Angles opposite to equal sides are equal)

In △ABD,

By angle sum property of triangle,

⇒ ∠DBA + ∠BAD + ∠BDA = 180°

⇒ x° + ∠BAC + ∠CAD + ∠BDA = 180°

⇒ x° + 65° + 30° + 30° = 180°

⇒ x° + 125° = 180°

⇒ x° = 180° - 125°

⇒ x° = 55°

⇒ x = 55.

Hence, the value of x = 55.

(iii) In △ABC,

AB = AC

⇒ ∠ABC = ∠ACB = 55° (Angles opposite to equal sides in a triangle are equal)

In △ABD,

By angle sum property of triangle,

⇒ ∠ABD + ∠BDA + ∠BAD = 180°

⇒ 55° + 75° + ∠BAD = 180°

⇒ 130° + ∠BAD = 180°

⇒ ∠BAD = 180° - 130°

⇒ ∠BAD = 50°.

In △ABC,

By angle sum property of triangle,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 55° + 55° + ∠BAD + ∠CAD = 180°

⇒ 55° + 55° + 50° + x° = 180°

⇒ 160° + x° = 180°

⇒ x° = 180° - 160°

⇒ x° = 20°

⇒ x = 20.

Hence, value of x = 20.