Question

Find: (i) $\angle B$

(ii) $\angle EDC$

(iii) $\angle ADE$

Answer 1

i) From $\Delta BCD$ we have, $\angle BCD={25}^o$ and $\angle BDC={100}^o$, then $\angle B={180}^o-({100}^o+{25}^o)={55}^o$ [Since the sum of angles of a triangle is ${180}^o$]

ii) Again $DE\parallel BC$ and $DC$ intersector then $\angle EDC=\angle BCD={25}^o$ [ alternate angle].

III) We have $\angle ADB={180}^o$, and $\angle BDC={100}^o$ and $\angle EDC={25}^o$.

Then $\angle ADE={180}^0-({100}^o+{25}^o)={55}^o$.