Solution:
$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ\dots(1)$
Now,
$\angle\text{A}+\angle\text{C}=2(\angle\text{B}+\angle\text{D})$ (given) ...(2)Also,
$\angle\text{A}=140^\circ,\ \angle\text{D}=60^\circ$Putting value of
$(\angle\text{A}+\angle\text{C})$ from eq. (2) in eq. (1)$2(\angle\text{B}+\angle\text{D})+\angle\text{B}+\angle\text{D}=360^\circ$
$3(\angle\text{B}+\angle\text{D})=360^\circ$
$\Rightarrow\angle\text{B}+\angle\text{D}=120^\circ$
$\Rightarrow\angle\text{B}+60^\circ=120^\circ$
$\Rightarrow\angle\text{B}=60^\circ$
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In the class intervals 10–20, 20–30, the number 20 is included in: