Question

LMN is an isosceles triangle. Find x",y
and z" in the figure given below. L =58​

Answer 1

$\large {\red {\mathfrak{☯ \: Given :}}}$

Given that, ∆LMN is an isosceles triangle and <NLM = 58°

$\large {\red {\mathfrak{☯ \: To \: find:}}}$

We have to find the values of x°, y° and z° in the triangle.

$\large {\red {\mathfrak{☯ \: Solution:}}}$

Since, ∆LMN is an isosceles triangle, <LNM = <LMN (Angles opposite to equal sides are equal).

Let, <LNM and <LMN be a.

So, by using angle sum property of triangle :

$\sf{ \: \: \: : \implies \angle LMN + \angle LNM +\angle MLN=180\degree}$

$\sf{ \: \: \: : \implies a + a + 58\degree=180\degree}$

$\sf \: \: \: :\implies 2a+58\degree=180\degree$

$\sf \: \: \: :\implies 2a=180\degree-58\degree$

$\sf \: \: \: :\implies2a=122\degree$

$\sf \: \: \: :\implies a= \cfrac{122\degree}{2}$

$\sf ~~~:\implies a=61 \degree$

$\boxed {\sf {\therefore \: \angle LMN(y\degree)=\angle LNM=61\degree}}$

Now, finding x° and z° because they make linear pair :

Here, z° = x° because <LMN = <LNM (If angle are equal, then their linear pairs will also be equal).

$\sf \: \: \: :\implies\angle LMN + z\degree=180\degree$

$\: \: \: :\implies \sf 61\degree+z\degree=180\degree$

$\sf \: \: \: :\implies z\degree=180\degree - 61 \degree$

$\: \: \: :\implies \sf z\degree=119\degree$

$\boxed{\sf \therefore \: z\degree=x\degree=119\degree}$