# 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 + 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 + 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,

x = 20° + 20°

x = 40°

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