Q. 17

Calculate the value of x in each of the following figure.

(i).

(ii)

(iii)

(iv)

(v)

(vi)

Answer :

(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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