Q. 17

# Calculate the value of x in each of the following figure.(i). (ii) (iii) (iv) (v) (vi) (i) 5

Given, BAE = 110° and ACD = 120°

ACB + ACD = 180° [Because BD is a straight line]

ACB + 120° = 180°

ACB = 60°_______________ (i)

In triangle ABC,

BAE = ABC + ACB

110° = x + 60°

x = 5

(ii) 120°

In triangle ABC,

A + B + C = 180° [Sum of angles of triangle ABC]

30° + 40° + C = 180°

C = 110°

BCA + DCA = 180° [Because BD is a straight line]

110° + DCA = 180°

DCA = 70°_________________ (i)

In triangle ECD,

AED = ECD + EDC

x = 70°+ 50°

x = 120°

(iii) 55°

Explanation:

BAC = EAF = 60°[Opposite angles]

In triangle ABC,

ABC + BAC = ACD

+ 60°= 115°

= 55°

(iv) 75°

Given AB||CD

Therefore,

BAD = EDC = 60°[Alternate angles]

In triangle CED,

C + D + E = 180°[Sum of angles of triangle]

45° + 60° + x = 180°[EDC = 60°]

x = 75°

(v) 30°

Explanation:

In triangle ABC,

BAC + BCA + ABC = 180°[Sum of angles of triangle]

40° + 90° + ABC = 180°

ABC = 50°________________ (i)

In triangle BDE,

BDE + BED + EBD = 180°[Sum of angles of triangle]

x° + 100° + 50° = 180°[EBD = ABC = 50°]

x° = 30°

(vi) x=30

Explanation:

In triangle ABE,

BAE + BEA + ABE = 180°[Sum of angles of triangle]

75° + BEA + 65° = 180°

BEA = 40°

BEA = CED = 40°[Opposite angles]

In triangle CDE,

CDE + CED + ECD = 180°[Sum of angles of triangle]

x° + 40° + 110° = 180°

x° = 30°

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