Question

In the given figure, $AB,BC$ and $AC$ touch the circle at $L.M$ and $N$ respectively. If $\angle B={70}^o$ and $\angle C={60}^o$, then the measure of $\angle LON$ is

• A. ${50}^o$
• B. ${110}^o$
• C. ${120}^o$
• D. ${130}^o$

Answer 1

The correct option is D ${130}^o$

Given that:

$\angle B={70}^{\circ },\angle C={60}^{\circ }$

To find:

$\angle LON=?$

Solution:

In $△ABC$

$\angle A+\angle B+\angle C={180}^{\circ }$

or, $\angle A+{70}^{\circ }+{60}^{\circ }={180}^{\circ }$

or, $\angle A={180}^{\circ }-{130}^{\circ }$

or, $\angle A={50}^{\circ }$

In quadrilateral $LOMB$

$\angle LOM={360}^{\circ }-{70}^{\circ }-2\times {90}^{\circ }$

$\angle LOM={360}^{\circ }-{70}^{\circ }-{180}^{\circ }$

$\angle LOM={110}^{\circ }$

Similarly

In quadrilateral $NOMC$

$\angle NOM={360}^{\circ }-{60}^{\circ }-2\times {90}^{\circ }$

$\angle NOM={360}^{\circ }-{60}^{\circ }-{180}^{\circ }$

$\angle NOM={120}^{\circ }$

Now,

$\angle LON+\angle LOM+\angle NOM={360}^{\circ }$ (angle at a point .)

$\angle LON+{110}^{\circ }+{120}^{\circ }={360}^{\circ }$

$\angle LON={360}^{\circ }-{230}^{\circ }$

$\angle LON={130}^{\circ }$

Hence, D is the correct option.