The sides of a right angled triangle are $21$ and $28 \; cm$. Find the length of the perpendicular drawn to the hypotenuse from the right angle.
Let $ABC$ be the given triangle with sides $AB = c = 28 \; cm$ and $AC = b = 21 \; cm$ and right angled at $A$.
Hypotenuse of $\triangle ABC$ $= AB = a = \sqrt{b^2 + c^2} = \sqrt{\left(21\right)^2 + \left(28\right)^2} = \sqrt{1225} = 35 \; cm$
In $\triangle ABC$, by sine rule we have $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B}$
i.e. $\;$ $\dfrac{35}{\sin 90^{\circ}} = \dfrac{21}{\sin B}$
i.e. $\;$ $\sin B = \dfrac{21}{35} = \dfrac{3}{5}$ $\;\;\; \cdots \; (1)$
Let $AP$ be the perpendicular from the right angle to the hypotenuse.
In $\triangle APB$, by sine rule we have $\;$ $\dfrac{AB}{\sin P} = \dfrac{AP}{\sin B}$
i.e. $\;$ $\dfrac{28}{\sin 90^{\circ}} = \dfrac{AP}{3 / 5}$ $\;\;\;$ [by equation $(1)$]
i.e. $\;$ $AP = 28 \times \dfrac{3}{5} = \dfrac{84}{5} = 16.8 \; cm$