In a Delta ABC,if the median,bisector,and alt | vedclass

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(C) Let the angle at vertex $A$ be divided into four equal parts of $	heta$ each. Thus,$\angle A = 4	heta$.From the figure,the altitude $AN$ makes $

Vedclass · Accepted Answer

$\frac{\pi}{2}, \frac{3\pi}{8}, \frac{\pi}{8}$

Vedclass · Answer

(C) Let the angle at vertex $A$ be divided into four equal parts of $	heta$ each. Thus,$\angle A = 4	heta$.From the figure,the altitude $AN$ makes $\angle BAN = \frac{\pi}{2} - B = 	heta$,so $B = \frac{\pi}{2} - 	heta$.Similarly,$\angle CAM = \frac{\pi}{2} - C = 	heta$,so $C = \frac{\pi}{2} - 3	heta$ (since $\angle MAC = 	heta$ and the total angle is $4	heta$).In $\Delta ABM$,by the sine rule: $\frac{BM}{\sin 	heta} = \frac{AB}{\sin(\pi - (B+	heta))} = \frac{AB}{\sin(B+	heta)}$.Since $BM = a$ and $AB = \frac{AN}{\cos 	heta}$,we have $\frac{a}{\sin 	heta} = \frac{AN}{\cos 	heta \sin(\pi/2)} = \frac{AN}{\cos 	heta}$.In $\Delta ANC$,$\frac{NC}{\sin 	heta} = \frac{AC}{\sin(\pi/2)} = AC$. Using the ratio of areas or sine rule,we find $	heta = \frac{\pi}{8}$.Thus,$A = 4	heta = \frac{\pi}{2}$,$B = \frac{\pi}{2} - \frac{\pi}{8} = \frac{3\pi}{8}$,and $C = \frac{\pi}{2} - \frac{3\pi}{8} = \frac{\pi}{8}$.The angles are $\frac{\pi}{2}, \frac{3\pi}{8}, \frac{\pi}{8}$.

In a $\Delta ABC$,if the median,bisector,and altitude drawn from the vertex $A$ divide the angle at the vertex into four equal parts,then the angles of the $\Delta ABC$ are:

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