Question

In fig.6.29, find <ACD and <AED.​

Answer 1

/_ AED=/_B+/_E (exterior /_ = sum of two interior angles)

=30°+30°

=60°

Now,

/_ E+/_C+/_D=180°

2/_C+50°=180°

2/_C=180°-50°

2/_C=130°

/_C=130°/2

/_C=65°

/_E=/_C=65°

/_C=65°

/_E=60°

Answer 2

Answer:

Step-by-step explanation:in ∆ABC

angleA+angleB+angleC=180

30+30+x=180

x=180-60

x=120

<ACB+<ECD=180(linear pair)

120+<ECD=180

<ECD=180-120=60

In ∆ ECD

<E+<C+<D=180

60+50+<E=180

<E=70

<AED+<CED=180

<AED=180-70=110