Q. 34.8( 31 Votes )

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


Answer :

(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°


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