$$ABC$$ is a triangle, $$\angle B=60^\circ, \angle C=45^\circ$$, BC is produced to / extended till D so that $$\angle ADB=30^\circ$$, then given $$\sqrt{3}=\frac{19}{11}$$ and $$BC \times CD=3^{4}\times 2^{3}\times\frac{11}{19}$$, what will be the square of the altitude from A to BC?
AE is altitude from A on BC. $$\angle\ AEC\ =\ 90^{\circ\ }$$
It is given,
$$\angle\ ABC\ =\ 60^{\circ\ },\ \angle\ ACB\ =\ 45^{\circ\ }$$
In triangle ABE,
AE =Â $$\sqrt{\ 3}x$$
As $$\angle\ EAC\ =\angle\ ECA$$, EC =Â $$\sqrt{\ 3}x$$
In triangle AED, $$\angle\ EAD\ =\ 60^{\circ\ }$$
Therefore, CD =Â $$3x-\sqrt{\ 3}x$$
It is given,
$$BC\ \cdot\ CD\ =\ \frac{3^4.2^3.11}{19}$$
$$x\left(\sqrt{\ 3}+1\right)\sqrt{\ 3}\left(\sqrt{\ 3}-1\right)x\ =\ \frac{3^4.2^3.11}{19}$$
$$x\ =\ \frac{3^2.2.11}{19}$$
AE = 9*2 = 18
$$AE^2$$ = 324
Answer is option B.
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