Question

In the given figure $OBC$ and $OKH$ are straight lines. If $AH=AK,b={80}^{\circ }andc={30}^{\circ }$ then the value of $d$ is

• A. ${20}^{\circ }$
• B. ${25}^{\circ }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is B ${25}^{\circ }$

In $△ABC$

$\angle ACB={30}^o$ and $\angle ABC={80}^o.$

$\Rightarrow$ $\angle BAC+\angle ACB+\angle ABC={180}^o$

$\Rightarrow$ $\angle BAC+{30}^o+{80}^o={180}^o$

$\Rightarrow$ $\angle BAC+{110}^o={180}^o$

$\therefore$ $\angle BAC={70}^o$

$\therefore$ $\angle KAH={70}^o$

In $△AHK$,

$AH=AK$ [ Given ]

$\angle AHK=\angle AKH$ [ Base angles of equal sides are also equal ] ---- ( 1 )

$\Rightarrow$ $\angle KAH+\angle AHK+\angle AKH={180}^o.$

$\Rightarrow$ ${70}^o+2\angle AKH={180}^o.$

$\Rightarrow$ $2\angle AKH={110}^o$

$\therefore$ $\angle AKH={55}^o.$

$\Rightarrow$ $\angle AKH=\angle BKO={55}^o.$ [ Vertically opposite angles ] ---- ( 2 )

$\Rightarrow$ $\angle ABC+\angle ABO={180}^o$ [ Linear pair ]

$\Rightarrow$ ${80}^o+\angle ABO={180}^o$

$\therefore$ $\angle ABO={100}^o$

$\therefore$ $\angle KBO={100}^o$ ---- ( 3 )

In $△BOK,$

$\Rightarrow$ $\angle BOK+\angle BKO+\angle KBO={180}^o$

$\Rightarrow$ $d+{55}^o+{100}^o={180}^o$ [ From ( 2 ) and ( 3 ) ]

$\Rightarrow$ $d+{155}^o={180}^o$

$\therefore$ $d={25}^o$