Points $D, E$ are taken on the side $B C$ of a triangle $A B C$ , such that $B D = D E = E C,$ if $∠ B A D = x, ∠ D A E = y, ∠ E A C = z$ then the value of $\frac{sin (x + y) sin (y + z)}{sin x sin z}$ is equal to

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Solution

We have, $B D = D E = E C = \frac{1}{3} a$
$ln △ A B E, \frac{sin (x + y)}{\frac{2}{3} a} = \frac{sin B}{A E}$ ...(1)

$ln △ A D C, \frac{sin (y + z)}{\frac{2}{3} a} = \frac{sin C}{A D}$ ...(2)

$ln △ A B D, \frac{sin x}{\frac{1}{3} a} = \frac{sin B}{A D}$ ...(3)

$ln △ A F C, \frac{sin z}{\frac{1}{3} a} = \frac{sin C}{A E}$ ...(4)

From (1), (2), (3), and (4), we have
$\frac{sin (x + y) sin (y + z)}{sin x sin z} = 4.$

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Points D,E are taken on the side BC of a triangle ABC, such that BD=DE=EC, if ∠BAD=x,∠DAE=y,∠EAC=z then the value of sin(x+y)sin(y+z)sinxsinz is equal to

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Points $D, E$ are taken on the side $B C$ of a triangle $A B C$ , such that $B D = D E = E C,$ if $∠ B A D = x, ∠ D A E = y, ∠ E A C = z$ then the value of $\frac{sin (x + y) sin (y + z)}{sin x sin z}$ is equal to