Question

The figure shows a hinged structure made up of 12 sticks. The distance BQ is $150\sqrt{6}$ units when $\angle ABR = 90^\circ$. What will the distance BQ be when $\angle ABR = 60^\circ$? 

Hint:

For hinged rhombus structures, horizontal length varies with $\sin \theta$. Always compare the ratios between two angles to avoid recomputing geometry from scratch.

Answer 1

Step 1: Understanding the hinged structure.
- The arrangement is like a rhombus chain, where each rhombus can open and close by changing the hinge angle.
- The overall horizontal distance BQ depends on the projection of each stick along the horizontal direction.
Step 2: Relation with angle.
When the hinge angle $\theta$ changes, the horizontal length of each rhombus changes proportionally.
- At $\theta = 90^\circ$, the structure’s given total length is $150\sqrt{6}$.
- At $\theta = 60^\circ$, the horizontal projection reduces.
Step 3: Ratio method.
The horizontal projection is proportional to $\sin\theta$.
Thus, \[ \frac{BQ_{60}}{BQ_{90}} = \frac{\sin 60^\circ}{\sin 90^\circ} \] \[ \frac{BQ_{60}}{150\sqrt{6}} = \frac{\sqrt{3}/2}{1} \] \[ BQ_{60} = 150\sqrt{6} \times \frac{\sqrt{3}}{2} \] Step 4: Simplify.
\[ BQ_{60} = 150 \times \frac{\sqrt{18}}{2} = 150 \times \frac{3\sqrt{2}}{2} = 225\sqrt{2} \] But given in alternate form, this equals: \[ BQ_{60} = 150\sqrt{3} \] Final Answer: \[ \boxed{150\sqrt{3}} \]