Question

The sum of two opposite angles of a parallelogram is ${130}^o$. Find the measure of each of its angles

• A. ${65}^o,{115}^o,{65}^o,{115}^o$
• B. ${65}^o,{105}^o,{65}^o,{115}^o$
• C. ${65}^o,{115}^o,{55}^o,{115}^o$
• D. ${65}^o,{115}^o,{65}^o,{105}^o$

Answer 1

The correct option is A ${65}^o,{115}^o,{65}^o,{115}^o$

$LetABCDaparallelogramwithfouranglesas\angle A,\angle B,\angle Cand\angle D\angle A+\angle C=130\left(Given\right)Also\angle A=\angle C(Oppositeanglesof\parallel gramareequal).......\left(1\right)⟹\angle A+\angle A=130⟹2\angle A=130⟹\angle A=\frac{130}{2}=65⟹\angle A=\angle C=\mathrm{65....}\left(2\right)Also\angle A+\angle B=180\left(Sumofcointerioranglesis180\right)⟹65+\angle B=180⟹\angle B=180-65=115Also\angle B=\angle D=115(Oppositeanglesof\parallel gramareequal)Hencealltheanglesare65,115,65and115$