Question

2. In the adjoining figure, ABCD is a rhombus in which ZBAC = 40°.
Find:
(a) ZACB
(b) ZABC
(c) ZADC
(d) ZACD
(e) ZCAD

Z = angle

Give me proper answer with explanation
plz​

Answer 1

Step-by-step explanation:

in fig.ABCD

given angle bac=40'

(a)so in triangle ACb its an isocles triangle so base angle are equal so angle ABC is 40'

now,

angle bac +angle ABC+angle BCA =180'(angle sum property)

40'+40'+angle BCA =180'

80'+angle BCA =180'

angle BCA=180-80=100'

so,angle BCA=100'

(b)angle abc=40'(given in a part)

(c)as its square so all angles measure 90'

so at angle a 90-40=50'

so,angle cad =50'

now,angle a =angle c(alternate angles)

anglec=40'

now angle DAC +angle DCA +angleCDA=180'(angle sum property)

50+40+angle CDA=180'

90+angle CDA=180'

angle CDA=180-90=90'

(d)angle ACD=40'given in c part

(e) angle cad =50'given in c part

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