Question

solve 21(i) and 21(ii)

Answer 1

21a)

in cyclic quadrilateral ABCD
I) ADC = CBE [exterior angle= opposite interior angle]
ADC = 100

ii)DAC = 1/2 (DOC)
[angle at center = 1/2(angle at circumference)]
DAC =1/2 x 40=20

iii)OD= OC[radius of the circle is equal]
OCD= ODC = x
in triangle ODC
x + x + 40=180
x = 70
ODC = 70
Now
ODA
= ADC-ODC
= 100-70
=30

iv)in triangle ADC
100 + 20 + ACD = 180
ACD = 60
Now
OCA = OCD-ACD= 70-60=10
_______________________________

21b)

in triangle ABC
55 + 65+BAC =180
BAC = 60
Now
BAD
= DAC
=1/2 BAC
[I is the in center and AD IS inspector]
=1/2 (60)
=30°

I)BCD
= BAD [angle at same segment]
=30

ii)CBD= DAC=30
Now
BCI = 1/2 ACB
[I Is the in center ]
= 1/2x65
=32.5

iii)DCI = BCI +BCD
=32.5 +30
=62.5
Now
IBC = 1/2 (ABC)
=1/2 x 55
=27.5

iv)in triangle CBI
27.5 + 32.5 + BIC = 180
BIC = 120