Question

Find the value of the unknown interior angle $x$ in the following figures:

Answer 1

For fig $\left(i\right)$

$\angle BCA+\angle ACD={180}^o$ [ Linear pair ]

$\angle BCA+{115}^o={180}^o$

$\therefore$ $\angle BCA={180}^o-{115}^o={65}^o$

In $△ABC$,

$\Rightarrow$ $\angle ABC+\angle BCA+\angle CAB={180}^o$

$\Rightarrow$ $x+{65}^o+{50}^o={180}^o$

$\Rightarrow$ $x+{115}^o={180}^o$

$\therefore$ $x={180}^o-{115}^o={65}^o$

For fig $\left(ii\right)$

$\angle PRQ+\angle QRS={180}^o$ [ Linear pair ]

$\angle PRQ+{100}^o={180}^o$

$\therefore$ $\angle PRQ={180}^o-{100}^o={80}^o$

In $△PQR,$

$\Rightarrow$ $\angle PRQ+\angle RQP+\angle QPR={180}^o$

$\Rightarrow$ ${80}^o+x+{70}^o={180}^o$

$\Rightarrow$ ${150}^o+x={180}^o$

$\therefore$ $x={180}^o-{150}^o={30}^o$

For fig $\left(iii\right)$,

$\angle NLM+\angle KLN={180}^o$

$\angle NLM+{1250}^o={180}^o$

$\therefore$ $\angle NLM={180}^o-{125}^o={55}^o$

In $△LMN,$

$\Rightarrow$ $\angle LMN+\angle MNL+\angle NLM={180}^o$

$\Rightarrow$ ${90}^o+x+{55}^o={180}^o$

$\Rightarrow$ ${145}^o+x={180}^o$

$\therefore$ $x={180}^o-{145}^o={35}^o$