Mathematics

In the adjoin figure, $$ABC$$ is an isocules triangle with $$AB=AC$$ & $$LM$$ is parallel to $$BC$$. If $$\angle A=50^o$$, find $$\angle LMC$$.


SOLUTION
REF.Image.
Given $$ \triangle ABC $$ is isosceles $$ \triangle $$ with $$ AB = AC $$
$$ LM \parallel BC $$ & $$ \angle A = 50^{\circ}$$ 
If $$ \triangle ABC $$ is isosceles then $$ \angle B = \angle C $$
$$ \angle A + \angle B + \angle C = 180^{\circ}$$
$$ 50^{\circ} + 2\angle B = 180 $$
$$ 2\angle B = 180 -50^{\circ}$$
$$ \angle B = \dfrac{130}{2}$$
$$ \angle B = 65^{\circ} = \angle C $$
Now $$ \angle MCB  = \angle AML (\therefore $$ cooreponding $$ \angle $$ b/w two || lines)
$$ 50^{\circ} \Rightarrow  65^{\circ} = \angle AML ...(1)$$
Now at point $$M \angle AMC = 180^{\circ}$$
$$ \angle AML + \angle LMC = 180^{\circ}$$  $$ (\therefore \angle AMC = \angle AML +\angle LMC)$$
from (1)
$$ 65^{\circ}+ \angle LMC = 180^{\circ}$$
$$ \angle LMC = 180^{\circ}-65^{\circ}$$
$$ \boxed{\angle LMC = 115^{\circ}}$$
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