Question

If the line $x=y=z$ intersect line in $\sin A+\sin B+\sin C={2d}^2$, $\sin 2A+\sin 2B+\sin 2C=d^2$Then $\sin \frac{A}{2}\sin \frac{B}{2}\frac{C}{2}$ is equal to

• A. $\frac{1}{16}$
• B. $\frac{1}{4}$
• C. $\frac{1}{13}$
• D. $\frac{1}{15}$

Answer 1

The correct option is A $\frac{1}{16}$

Given line is $x=y=z...\left(1\right)$

which intersects given two lines are

$\left(sinA\right)x+\left(sinB\right)y+\left(sinC\right)z=2d^2...\left(2\right)$

and, $\left(sin2A\right)x+\left(sin2B\right)y+\left(sin2C\right)z=d^2...\left(3\right)$

Let the intersection point for $\left(1\right)$ and $\left(2\right)$ is $(k,k,k)$.

$\therefore$ The equation $\left(2\right)$ becomes

$\left(sinA\right)k+\left(sinB\right)k+\left(sinC\right)k=2d^2$

$\Rightarrow k=\frac{2d^2}{sinA+sinB+sinC}$

$\Rightarrow k=\frac{2d^2}{4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}$

$\Rightarrow k=\frac{d^2}{2cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}...\left(4\right)$

Similarly, for equation $\left(1\right)$ and equation $\left(3\right)$, we obtained the value of

$k=\frac{d^2}{4\left(2sin\frac{A}{2}cos\frac{A}{2}\right)\left(2sin\frac{B}{2}cos\frac{B}{2}\right)\left(2sin\frac{C}{2}cos\frac{C}{2}\right)}$

$\Rightarrow k=\frac{d^2}{32\left(sin\frac{A}{2}cos\frac{A}{2}\right)\left(sin\frac{B}{2}cos\frac{B}{2}\right)\left(sin\frac{C}{2}cos\frac{C}{2}\right)}...\left(5\right)$

Equating equation $\left(4\right)$ and $\left(5\right)$, we have

$\frac{d^2}{2cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}}$ $=\frac{d^2}{32\left(sin\frac{A}{2}cos\frac{A}{2}\right)\left(sin\frac{B}{2}cos\frac{B}{2}\right)\left(sin\frac{C}{2}cos\frac{C}{2}\right)}$

$\Rightarrow sin\frac{A}{2}sin\frac{B}{2}sin\frac{C}{2}=\frac{1}{16}$

So, $\text{A}$ is the correct option.