Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the are AC such that , and , then is equal to [2025]

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Solution

Q.
Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the are AC such that $\frac{length of arc A B}{length of arc B C} = \frac{1}{5}$ , and $\vec{O C} = α \vec{O A} + β \vec{O B}$ , then $α + \sqrt{2} (\sqrt{3} - 1) β$ is equal to [2025]

1 $2 - \sqrt{3}$

2 $5 \sqrt{3}$

3 $2 \sqrt{3}$

4 $2 + \sqrt{3}$

Ans.
(1)

Here, $∠ A O B = 15 °$ and $∠ B O C = 75 °$

Given $\vec{O C} = α \vec{O A} + β \vec{O B}$ ... (i)

$\vec{O A} \cdot \vec{O C} = α \vec{O A} \cdot \vec{O A} + β \vec{O B} \cdot \vec{O A}$

$0 = α | \vec{O A} |^{2} + β$ $| \vec{O B} |$ $| \vec{O A} |$ $\cdot \cos 15 °$ [ $∵ \vec{O A} ⊥ \vec{O C}$ ]

$\Rightarrow α + β \cos 15 ° = 0$ ... (ii)

[ $∵$ $| \vec{O A} |$ $=$ $| \vec{O B} |$ $=$ $| \vec{O C} |$ ]

$\vec{O B} \cdot \vec{O C} = α \vec{O A} \cdot \vec{O B} + β \vec{O B} \cdot \vec{O B}$ [Form (i)]

$| \vec{O B} |$ $| \vec{O C} |$ $\cos 75 ° = α$ $| \vec{O A} |$ $| \vec{O B} |$ $\cos 15 ° + β | \vec{O B} |^{2}$

$\Rightarrow \cos 75 ° = α \cos 15 ° + β$ ... (iii)

From (ii) and (iii), we get

$\cos 75 ° = - β \cos^{2} 15 + β$

$\Rightarrow \cos 75 ° = β (1 - \cos^{2} 15 °) = β \sin^{2} 15 °$

$β = \frac{\sin 15 °}{\sin^{2} 15 °} = \frac{1}{\sin 15 °} = \frac{2 \sqrt{2}}{\sqrt{3} - 1}$ [ $∵ \cos (90 ° - θ) = \sin θ$ ]

From (ii), $α = - β \cos 15 ° = \frac{- 2 \sqrt{2}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

$\Rightarrow α = \frac{- (\sqrt{3} + 1)}{(\sqrt{3} - 1)}$

$∴ α + \sqrt{2} (\sqrt{3} - 1) β = \frac{- (\sqrt{3} + 1)}{(\sqrt{3} - 1)} + \sqrt{2} \frac{(\sqrt{3} - 1) \times 2 \sqrt{2}}{(\sqrt{3} - 1)} = (2 - \sqrt{3})$

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