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Mano Kannan

· started a discussion

· 1 Months ago

quants 1 (Question:1096096.0)

how d=68

Question:

In a quadrilateral ABCD , if ∠A is three-fifth of the ∠B, ∠C is 45% more than ∠B. ∠D is 68˚ more than ∠A. find the difference between ∠C and ∠B.

Options:
A)

45o

B)

60o 

C)

34o 

D)

36o 

Solution:

Ans: (d) 

Given four angles of quadrilateral ABCD are ∠A, ∠B, ∠C, ∠D 

As we know, sum of all angles of a quadrilateral is 360˚ 

∠A + ∠B + ∠C + ∠D = 360˚ 

Let ∠B = x˚ 

3x/5 + x + 1.45x + 3x/5 + 68˚ = 360˚ 

By solving we get

x = 80˚ 

∠A = 48˚ ∠B =80˚ ∠C =116˚ ∠D = 116˚ 

the difference between ∠C and ∠B = 116˚-80˚= 36˚ 

Hence option d is correct.

Knowledge Expert

· commented

· 1 Months ago

Dear student,
Given answer is correct. D is 68 more than A and A is 3x/5 so D is 3x/5+68.

Regards
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