Question

In the adjoining figure, $$LM=MO$$ and $$LN=NM$$. Find: a) $\mathrm{\angle }NLM$ b) $\mathrm{\angle }LOM$.

Answer 1

Hint: We will use a concept that the angle opposite to the equal sides of the triangles are of equal measurement. For example, if ABC is a triangle, with $AB=AC$ then we can say that $\mathrm{\angle }ABC=\mathrm{\angle }ACB$. We will also use the angle sum property of triangles which state that the sum of all the angles of the triangle is equal to 180.

Complete step-by-step answer:
It is given in the question that $LM=MO$ and $LN=NM$ then we will have to find the angles $\mathrm{\angle }NLM$ and $\mathrm{\angle }LOM$.

We know that sum of all angles of a triangle is equal to 180 degree, so, in $\mathrm{\Delta }LNM$, we have $\mathrm{\angle }MNL+\mathrm{\angle }NLM+\mathrm{\angle }LMN={180}^{{}^{\circ }}.....(i)$.
We also know that angles opposite to the sides of equal length are of equal measurement value. In $\mathrm{\Delta }LMN$, we have side $$LN=NM$$, so from this, we can say that $\mathrm{\angle }NLM=\mathrm{\angle }LMN$.
On putting these value in the equation $\mathrm{\angle }MNL+\mathrm{\angle }NLM+\mathrm{\angle }LMN={180}^{{}^{\circ }}.....(i)$ we get,
${60}^{{}^{\circ }}+\mathrm{\angle }NLM+\mathrm{\angle }NLM={180}^{{}^{\circ }}$, solving further, we get,
$2\mathrm{\angle }NLM={180}^{{}^{\circ }}-{60}^{{}^{\circ }}$, therefore, we get finally,
$$\mathrm{\angle }NLM={\displaystyle \frac{{120}^{{}^{\circ }}}{2}}={60}^{{}^{\circ }}$$.
Also, we get $\mathrm{\angle }NML={60}^{{}^{\circ }}$.
We know that sum of linear pairs is equal to 180, and angles $\mathrm{\angle }LMOand\mathrm{\angle }LMN$ are forming linear pair, therefore,
$\mathrm{\angle }LMO+\mathrm{\angle }LMN={180}^{{}^{\circ }}$, putting the value $\mathrm{\angle }NML={60}^{{}^{\circ }}$, we get,
$\mathrm{\angle }LMO+{60}^{{}^{\circ }}={180}^{{}^{\circ }}$, therefore,
$\mathrm{\angle }LMO={180}^{{}^{\circ }}-{60}^{{}^{\circ }}={120}^{{}^{\circ }}$.
Now, in $\mathrm{\Delta }LMO$, we have sum of all three angles of the triangle as 180, it means $\mathrm{\angle }LOM+\mathrm{\angle }OML+\mathrm{\angle }MLO={180}^{{}^{\circ }}$. Also, $\mathrm{\angle }LOM=\mathrm{\angle }MLO$ as they are angles opposite to equal sides, therefore we get,
$\mathrm{\angle }LOM+{120}^{{}^{\circ }}+\mathrm{\angle }LOM={180}^{{}^{\circ }}$, solving further, we get
$2\mathrm{\angle }LOM={180}^{{}^{\circ }}-{120}^{{}^{\circ }}={60}^{{}^{\circ }}$, therefore, we finally get,
$\mathrm{\angle }LOM={\displaystyle \frac{{60}^{{}^{\circ }}}{2}}={30}^{{}^{\circ }}$.
Thus we get $$\mathrm{\angle }NLM={60}^{{}^{\circ }}$$ and $\mathrm{\angle }LOM={30}^{{}^{\circ }}$.

Note: Usually students do miss-calculation in adding the angles of the triangle and it is enough to waste our all effort to solve this question. Generally, these silly mistakes are not traced even while revision, thus it is recommended to do calculation steps without mistakes. Sometimes students might think that triangle LMN is an equilateral triangle just by looking at it and then compute the angle, but it is not necessarily true always.