Q16 of 100 Page 17

In a cyclic quadrilateral ABCD, if ∠A-∠C=60°, prove that the smaller of two is 60°.

WE have,

∠A - ∠C = 60o (i)


Since, ABCD is a cyclic quadrilateral


Then,


∠A + ∠C = 180o (ii)


Adding (i) and (ii), we get


∠A - ∠C + ∠A + ∠C = 60o + 180o


2 ∠A = 240o


∠A = 120o


Put value of ∠A in (ii), we get


120o + ∠C = 180o


∠C = 60o


More from this chapter

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14

In Fig. 16.186, O is the centre of the circle. If ∠CEA=30°, find the values of x, y and z.

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In Fig. 16.187, ∠BAD=78°, ∠DCF=x° and ∠DEF=y°. Find the values of x and y.

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In Fig. 16.188, ABCD is a cyclic quadrilateral. Find the value of x.

 18

ABCD is a cyclic quadrilateral in which:

(i) BC||AD, ∠ADC=110° and ∠BAC=50°. Find ∠DAC.


(ii) ∠DBC=80° and ∠BAC=40° Find ∠BCD.


(iii) ∠BCD=100° and ∠ABD=70°. Find ∠ADB.