Q2 of 29 Page 63

ABCD is a cyclic quadrilateral. Given that ∠ADB + ∠DAB = 120°and ∠ABC + ∠BDA = 145°. Find the value of ∠CDB

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°


Let BAD = x°


Hence BCD = 180–x


Let BDC = a°


Let DBC = z°; ABC = 60 + z


Let ADB = y°


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°

More from this chapter

All 29 →
15

In the figure at left, O is the centre of the circle, ∠AOC = 100° and side AB is produced to D.

Find (i) ∠CBD (ii) ∠ABC


 1

O is the centre of the circle. AB is the chord and D is mid-point of AB. If the length of CD is 2cm and the length of chord is12 cm, what is the radius of the circle

 3

In the given figure, AB is one of the diameters of the circle and OC is perpendicular to it through the center O. If AC is 7√2 cm, then what is the area of the circle in cm2

 4

In the given figure, AB is a diameter of the circle and points C and D are on the circumference such that ∠CAD = 30°and ∠CBA = 70° what is the measure of ACD?