Given: △AKM, KD ⊥ AM , AK = 6, KM = 10, m∠AKM = 93º Find: KD

Answer: The measure of the perpendicular KD is 8.72 unit .Step-by-step explanation:Given as :The triangle AKM  , with KD perpendicular to AMThe measure of side AK = 6 unitThe measure of side KM = 10 unitThe ∠ AKM = 93°Let The measure of side KD = x unitNow, ∵ KD ⊥ AM , KD divide the angle  ∠ AKM  equallySo,  ∠ AKD =  ∠ [tex]\dfrac{AKM}{2}[/tex]I.e  ∠ AKD =  ∠ [tex]\{93}{2}[/tex]∴  ∠ AKD = 46.5°Now, AgainCos Ф =  [tex]\dfrac{AK}{KD}[/tex]I.e Cos 46.5° = [tex]\dfrac{6}{KD}[/tex]I.e 0.688 = [tex]\dfrac{6}{KD}[/tex]∴ KD = [tex]\dfrac{6}{0.688}[/tex]I.e KD = 8.72 unitHence The measure of the perpendicular KD is 8.72 unit  Answer