[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
→ In triangle KML, P and N are the points on KK and KM respectively such that
[tex]\rm \longmapsto\:KN = c[/tex]
[tex]\rm \longmapsto\:NM = b[/tex]
[tex]\rm \longmapsto\:PN = x[/tex]
[tex]\rm \longmapsto\:LM = a[/tex]
[tex]\rm \longmapsto\: \angle \: KNP = 50 \degree[/tex]
[tex]\rm \longmapsto\: \angle \: LKM = 50 \degree[/tex]
Now,
[tex]\rm \longmapsto\: In \: \triangle \: LKM \: and \: \triangle \:KNP[/tex]
[tex] \purple{\rm \longmapsto\:\angle \:PKN \: = \: \angle \:PKN \: \: \: \: \: \{common \}}[/tex]
[tex] \purple{\rm \longmapsto\:\angle \:KNP \: = \: \angle \:KML \: \: \: \: \: \{each \: 50 \degree \}}[/tex]
[tex] \purple{\rm\implies \:\triangle \:LKM \: \sim \: \triangle \:PKN \: \: \: (AA \: similarity)}[/tex]
[tex]\bf\implies \:\dfrac{KN}{KM} = \dfrac{PN}{LM} [/tex]
[tex]\rm \longmapsto\:\dfrac{c}{b + c} = \dfrac{x}{a} [/tex]
[tex]\bf\implies \:x \: = \: \dfrac{ac}{b + c}[/tex]
[tex]\textsf{Hope this helps!!}\\[/tex]
