Question

In the given figure, $OBC$ and $OKH$ are straight lines. If $AH=AK,b={80}^{\circ }$ and $c={30}^{\circ }$ 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 $△ABC,\angle A+\angle B+\angle C={180}^{\circ }$

$\angle A={180}^{\circ }-{80}^{\circ }-{30}^{\circ }={70}^{\circ }$

Given,$AH=AK⟹\angle AHK=\angle AKH$

In $△ABC,\angle AHK+\angle AKH+\angle HAK={180}^{\circ }$

$2\times \angle AKH={180}^{\circ }-{70}^{\circ }$

$\angle AKH=\frac{{110}^{\circ }}{2}={55}^{\circ }$

$\angle AKH=\angle BKO={55}^{\circ }$ ...... (Vertically opposite angles)

Since $CBO$ is a straight line,

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

$\angle OBK={180}^{\circ }-{80}^{\circ }={100}^{\circ }$

In $△KOB,\angle KOB+\angle KBO+\angle OKB={180}^{\circ }$

$d+{100}^{\circ }+{55}^{\circ }={180}^{\circ }$

$\Rightarrow d={25}^{\circ }$