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# ABCD is a cyclic quadrilateral. Given that ∠ADB + ∠DAB = 120°and ∠ABC + ∠BDA = 145°. Find the value of ∠CDBA. 75°B. 115°C. 35°D. 45°

As ADB + DAB = 120°

In triangle ABD as sum of all angles = 180°

ABD = 180–120 = 60°

As ABCD is cyclic, opposite angles have sum of 180°

Hence BCD = 180–x

Let BDC = a°

Let DBC = z°; ABC = 60 + z

As sum of angles of triangle BDC = 180°

180–x + a + z = 180

a = x–z ...(1)

According to question,

x + y = 120 ..(2)

and

60 + z + y = 145...(3)

Subtracting (3) from (2)

We get

x + y–60–z–y = –25

x–z = 35

Equating from (1) we get

a = 35°

hence CDB = 35°

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