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JEE Main 2023
Properties of Triangle
Properties of Triangle
Medium

Question

If the line x=y=zx=y=z intersects the line xsin⁡A+ysin⁡B+zsin⁡C−18=0=xsin⁡2A+ysin⁡2B+zsin⁡2C−9x \sin A+y \sin B+z \sin C-18=0=x \sin 2 A+y \sin 2 B+z \sin 2 C-9, where A,B,CA, B, C are the angles of a triangle ABCA B C, then 80(sin⁡A2sin⁡B2sin⁡C2)80\left(\sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right) is equal to ______________.

Answer: 18

Solution

x=y=z=k( let )∴k(sin⁡A+sin⁡B+sin⁡C)=18⇒k(4cos⁡A2⋅cos⁡B2⋅cos⁡C2)=18k(sin⁡2A+sin⁡2B+sin⁡2C)=9⇒k(4sin⁡A⋅sin⁡B⋅sin⁡C)=9… (ii) (ii)/(i)8sin⁡A2⋅sin⁡B2⋅sin⁡C2=918⇒80sin⁡A2⋅sin⁡B2⋅sin⁡C2=5\begin{aligned} &x= y=z=k(\text { let }) \\\\ &\therefore k(\sin A+\sin B+\sin C)=18 \\\\ &\Rightarrow k\left(4 \cos \frac{A}{2} \cdot \cos \frac{B}{2} \cdot \cos \frac{C}{2}\right)=18 \\\\ & k(\sin 2 A+\sin 2 B+\sin 2 C)=9 \\\\ &\Rightarrow k(4 \sin A \cdot \sin B \cdot \sin C)=9 \ldots \text { (ii) } \\\\ & \text {(ii)} /\text {(i)} \\\\ & 8 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}=\frac{9}{18} \\\\ &\Rightarrow 80 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}=5 \end{aligned}

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