Compute the value of x in each of the following figures:

(i) Given:

As AB = AC (isosceles triangle)

So,

∠ ABC = ∠ ACB = 50°

As B, C and D lie on the same line.

So,

∠ ACB + ∠ ACD = 180°

⇒ 50° + x = 180°

⇒ x = 180° - 50°

⇒ x = 130°

(ii)

As B, A and D lie on the same line.

So,

∠ BAC + ∠ CAD = 180°

⇒ ∠ BAC + 130° = 180°

⇒ ∠ BAC = 180° - 130°

⇒ ∠ BAC = 50°

As we know, using theorem (2) i.e. if a side of triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ABE = ∠ BAC + ∠ ACB

⇒ 106° = 50° + x

⇒ x = 106° - 50°

⇒ x = 56°

(iii)

∠ BAC = ∠ EAF = 65° (vertically opposite angle)

As we know, using theorem (2) i.e. if a side of triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ACD = ∠ BAC + ∠ CBA

⇒ 100° = 65° + x

⇒ x = 100° - 65°

⇒ x = 35°

(iv)

As C, A and D lie on the same line.

So,

∠ CAB + ∠ BAD = 180°

⇒ ∠ CAB + 120° = 180°

⇒ ∠ CAB = 180° - 120°

⇒ ∠ CAB = 60°

As we know, using theorem (2) i.e. if a side of triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ACE = ∠ CAB + ∠ ABC

⇒ 112° = 60° + x

⇒ x = 112° - 60°

⇒ x = 52°

(v)

As AB = BC (isosceles triangle)

So,

∠ BAC = ∠ BCA = 20°

As we know, using theorem (2) i.e. if a side of the triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

Hence,

∠ ADB = ∠ BAC + ∠ BCA

⇒ x = 20° + 20°

⇒ x = 40°