Question

In a triangle ABC, medians AD and BE are drawn. If AD = 4, ∠DAB = π/6 and ∠ABE =π/3, then the area of the ΔABC is

(a) 8/3

(b) 16/3

(c) 32/3√3

(d) 64/3

Answer 1

Question ⤵️

$\cos( \frac{\pi}{7} ) \times \cos( \frac{2\pi}{7} ) \times \cos( \frac{4\pi}{7} )$

Solution ⤵️

$\cos( \frac{\pi}{7} ) \times \cos( \frac{2\pi}{7} ) \times \cos( \frac{4\pi}{7} )$

$\implies \cos(20. \frac{\pi}{7} ) \times \cos( {2}^{1}. \frac{\pi}{7} ) \times \cos( {2}^{2} . \frac{\pi}{7} )$

$\implies \frac{ \sin {2}^{3} ( \frac{\pi}{7} ) }{ {2}^{} . \sin \frac{\pi}{7} }$

$( \because \{ \cos(a). \cos(2a)... \cos( {2}^{n - 1}a ) = \frac{ \sin( {2}^{n} a) }{ {2}^{n} \sin a \: }$

$\frac{ \sin(8 \frac{\pi}{7} ) }{8. \sin( \frac{\pi}{7} ) } = \frac{ \sin(\pi + \frac{\pi}{7} ) }{8. \sin( \frac{\pi}{7} ) }$

$\frac{ \sin(8 \frac{\pi}{7} ) }{8. \sin( \frac{\pi}{7} ) } = \frac{ \sin(\pi + \frac{\pi}{7} ) }{8. \sin( \frac{\pi}{7} ) } = \frac{ - \sin( \frac{\pi}{7} ) }{8. \sin( \frac{\pi}{7} ) }$

$\implies \frac{ - 1}{8}$

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Answer 2

Answer:

c) is the answer dear

Step-by-step explanation:

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