A horizontal park is in the shape of a triangle OAB with AB = 16. A vertical lamp post OP is erected at the point O such that\( \begin{array}{l}\angle PAO = \angle PBO = 15^\circ ~\text{and}~ \angle PCO = 45^\circ,\end{array}\)where C is the midpoint of AB. Then (OP)² is equal to

Question

If \(\frac {3cos\ 2x+cos^32x}{cos^6x-sin^6x}=x^3-x^2+6\), then find sum of roots.

Answer 1

To solve the problem, we will use the given information and apply trigonometric concepts.

The triangle OAB is given with its base AB = 16 and angle conditions at the point of the lamp post OP.
The midpoint C of AB gives us AC = CB = 8.
We know: \(\angle PAO = \angle PBO = 15^\circ\) and \(\angle PCO = 45^\circ\).
Using the Law of Sines in triangle OPC:
- Applying the law: \(\frac{OP}{\sin(45^\circ)} = \frac{OC}{\sin(15^\circ)}\)
We have:
- \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\)
- \(\sin(15^\circ) = \sin(45^\circ - 30^\circ)= \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ)\)
- Using known values: \(\sin(15^\circ) = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}\)
The OC can be calculated using coordinates. Since C is the midpoint of AB, C = (8, 0, 0). Thus, OC = 8.
Substituting known quantities into the Law of Sines formula:
- \(OP = \frac{\sqrt{2}}{2} \times \frac{8}{\frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{8 \cdot \sqrt{2} \cdot 4}{2(\sqrt{6} - \sqrt{2})}\)
- Simplifying gives: \(OP = \frac{32 \cdot \sqrt{2}}{\sqrt{6} - \sqrt{2}}\)
Now, we find \( (OP)^2 \), the required value:
- From rationalizing the denominator, we get the square term: \(\frac{32^2 \cdot 2}{(\sqrt{6} - \sqrt{2})\cdot (\sqrt{6} + \sqrt{2})}\)
- \(\sqrt{6}^2 - \sqrt{2}^2 = 6 - 2 = 4\)
- Thus: \( (OP)^2 = \frac{256 \cdot \sqrt{3}}{4} \times \frac{1}{2-\sqrt{3}} = \frac{32}{\sqrt{3}}(2-\sqrt{3})\)

Thus, the correct answer is: \(\frac{32}{\sqrt{3}}(2-\sqrt{3})\).

Answer 2

To solve the problem, we will use the given information and apply trigonometric concepts.

The triangle OAB is given with its base AB = 16 and angle conditions at the point of the lamp post OP.
The midpoint C of AB gives us AC = CB = 8.
We know: \(\angle PAO = \angle PBO = 15^\circ\) and \(\angle PCO = 45^\circ\).
Using the Law of Sines in triangle OPC:
- Applying the law: \(\frac{OP}{\sin(45^\circ)} = \frac{OC}{\sin(15^\circ)}\)
We have:
- \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\)
- \(\sin(15^\circ) = \sin(45^\circ - 30^\circ)= \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ)\)
- Using known values: \(\sin(15^\circ) = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2}\cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}\)
The OC can be calculated using coordinates. Since C is the midpoint of AB, C = (8, 0, 0). Thus, OC = 8.
Substituting known quantities into the Law of Sines formula:
- \(OP = \frac{\sqrt{2}}{2} \times \frac{8}{\frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{8 \cdot \sqrt{2} \cdot 4}{2(\sqrt{6} - \sqrt{2})}\)
- Simplifying gives: \(OP = \frac{32 \cdot \sqrt{2}}{\sqrt{6} - \sqrt{2}}\)
Now, we find \( (OP)^2 \), the required value:
- From rationalizing the denominator, we get the square term: \(\frac{32^2 \cdot 2}{(\sqrt{6} - \sqrt{2})\cdot (\sqrt{6} + \sqrt{2})}\)
- \(\sqrt{6}^2 - \sqrt{2}^2 = 6 - 2 = 4\)
- Thus: \( (OP)^2 = \frac{256 \cdot \sqrt{3}}{4} \times \frac{1}{2-\sqrt{3}} = \frac{32}{\sqrt{3}}(2-\sqrt{3})\)

Thus, the correct answer is: \(\frac{32}{\sqrt{3}}(2-\sqrt{3})\).

The Correct Option is B