Question

gm
If BD is diagonal of a 118 ABCD and
ZC = 5aº, ZCBD = 3aº and ZBDC = 2aº then
find all four angles of 118 ABCD.
gm

Answer 1

Answer:

gm If BD is diagonal of a 118 ABCD and ZC = 5aº, ZCBD = 3aº and ZBDC = 2aº then find all four angles of 118 ABCD. gm.

Step-by-step explanation:

I Think

Answer 2

Answer:

all angles = 90 degree

Step-by-step explanation:

in triangle BDC

angle bdc + angle dcb + angle cbd = 180 degree [angle sum property of triangle ]

2a + 5a + 3a = 180 degree

10a = 180 degree

a = 18 degree

angle dcb = 5a = 90 degree

angle dab = angle dcb = 90 degree [opposite angles of parallelogram are equal ]

angle dcb + angle cba = 180 degree [Co interior angles ]

90 degree + angle cba = 180 degree

angle cba = 90 degree

angle cba = angle cda = 90 degree [opposite angles of parallelogram are equal ]

all angles = 90 degree