In a Delta ABC,angle A = 30^circ and a = 8 te | vedclass

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(C) The distance of the orthocentre $H$ from vertex $A$ in a triangle is given by the formula $AH = 2R \cos A$.From the Sine Rule,we know that $\frac{

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$8\sqrt{3} 	ext{ cm}$

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(C) The distance of the orthocentre $H$ from vertex $A$ in a triangle is given by the formula $AH = 2R \cos A$.From the Sine Rule,we know that $\frac{a}{\sin A} = 2R$,which implies $2R = \frac{a}{\sin A}$.Substituting this into the formula for $AH$,we get $AH = \frac{a}{\sin A} 	imes \cos A = a \cot A$.Given $a = 8 	ext{ cm}$ and $\angle A = 30^\circ$,we have $AH = 8 	imes \cot(30^\circ)$.Since $\cot(30^\circ) = \sqrt{3}$,we get $AH = 8\sqrt{3} 	ext{ cm}$.

In a $\Delta ABC$,$\angle A = 30^\circ$ and $a = 8 \text{ cm}$,then the distance of the orthocentre from vertex $A$ is equal to (with usual notations).

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