Question

In the given circle with centre ${}^{\prime }{O}^{\prime }$, the mid points of two equal chords $AB$ & $CD$ are $K$ & $L$, respectively. If $\angle$ $OLK=$ ${25}^{\circ }$, Then $\angle$ $LKB$ is equal to:

• A. ${125}^{\circ }$
• B. ${115}^{\circ }$
• C. ${105}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

Answer: (B)

${115}^{\circ }$

Given: $O$ is the center of the circle. $K$ and $L$ are mid points of chords $AB$ and $CD$, respectively.
$\therefore OK\perp AB$ and $OL\perp CD$. {The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord}
As $AB=CD$.
$\therefore$ $OK=OL$ ...(Equal chords are equidistant from centre)

So, $△$ $OKL$ is an isosceles triangle.
$\therefore \angle OKL=\angle OLK={25}^{\circ }$

Therefore, $\angle LKB=\angle OKL+\angle OKB$

$={25}^{\circ }+{90}^{\circ }={115}^{\circ }$