Applying sine/cosine rule in non right triangle

Question

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

By applying first cosine, then sine rule I was able to find KD=~5.03 Can you confirm if this answer is correct? I repeated several times and got the same result. Thanks

Answer 1

$$ \frac{1}{2}\cdot |AK|\cdot |KM| \cdot \sin(\angle AKM) = \frac{1}{2} \cdot |AM|\cdot |KD|\\ |KD| = \frac{|AK| \cdot |KM| \cdot \sin(\angle AKM)}{|AM|}\\ |AM| = \sqrt{|AK|^2+|KM|^2-2\cdot |AK| \cdot |KM| \cdot \cos(\angle AKM)}\\ \therefore\space, |KD| = \frac{|AK| \cdot |KM| \cdot \sin(\angle AKM)}{\sqrt{|AK|^2+|KM|^2-2\cdot |AK| \cdot |KM| \cdot \cos(\angle AKM)}}\\ \space \\ $$ Substituting the respective values, we get $5.023$, to 3 decimal places.

Answer 2

$Area=$ $\;$$\dfrac{1}{2}|AK||KM|sin∠AKM=\dfrac{1}{2}|AM||KD|$ $$\\$$ $|AK|$,$\;$ $|KM|$ are given. Also by The Law of Cosines, you can find $|AM|$, thus you can find $|KD|$.