Question

$ABCD$ is a parallelogram in which $\angle DAB={65}^{\circ }$and $\angle CBD={65}^{\circ }$Find $\angle CDB$and $\angle ADB$

Answer 1

Step-1: Find $\angle ABD$

$\angle DAB+\angle CBA={180}^{\circ }$(Adjacent angles of a parallelogram are equal) ------ (i)

$$\begin{gathered} \angle CBA=\angle CBD+\angle ABD \\ \angle CBA={50}^{\circ }+\angle ABD \end{gathered}$$ (Given) ---------- (ii)

$\angle DAB+\angle CBA={180}^{\circ }$

$$\begin{gathered} \angle DAB+\angle ABD+{50}^{\circ }={180}^{\circ } \\ {65}^{\circ }+{50}^{\circ }+\angle ABD={180}^{\circ } \\ {115}^{\circ }+\angle ABD={180}^{\circ } \\ \angle ABD={180}^{\circ }-{115}^{\circ } \\ \angle ABD={65}^{\circ } \end{gathered}$$from (i) and (ii)

Step-2: Find $\angle CDB$

$\angle CDB$$=\angle ABD$(Alternate interior angle)

$\angle CDB$$={65}^{\circ }$

Step-3 Find $\angle ADB$

In $∆ADB$

$\angle DAB+\angle ABD+\angle ADB={180}^{\circ }$(Sum of angles of a triangle is ${180}^{\circ }$

$$\begin{gathered} {65}^{\circ }+{65}^{\circ }+\angle ADB={180}^{\circ } \\ {130}^{\circ }+\angle ADB={180}^{\circ } \\ \angle ADB={180}^{\circ }-{130}^{\circ } \\ \angle ADB={50}^{\circ } \end{gathered}$$

Hence, angles $\angle CDB={65}^{\circ }and\angle ADB={50}^{\circ }$