Question

The value of $\theta$ lies in the interval

• A. (0, 15°)
• B. (15°, 30°)
• C. 30°, 45°)
• D. (45°, 60°)

Answer 1

Answer: (D)

(D) (45°, 60°)

Answer 2

The circle is $x^2 + y^2 = 4$, centered at O(0, 0) with radius $r = 2$. The point is P(4, 2). The distance $OP = \sqrt{4^2 + 2^2} = \sqrt{20} = 2\sqrt{5}$. Let $\theta$ be the angle between the tangents from P, and let $\alpha$ be half of this angle ($\theta = 2\alpha$). In the right-angled triangle OAP (where A is a point of contact), $\sin \alpha = \frac{OA}{OP} = \frac{2}{2\sqrt{5}} = \frac{1}{\sqrt{5}}$. From $\sin \alpha = 1/\sqrt{5}$, we can find $\tan \alpha = \frac{1}{2}$. Using the double angle formula for tangent, $\tan \theta = \tan(2\alpha) = \frac{2 \tan \alpha}{1 - \tan^2 \alpha} = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$. Since $\tan 45^\circ = 1$ and $\tan 60^\circ = \sqrt{3} \approx 1.732$, and $1 < 4/3 < \sqrt{3}$, the angle $\theta$ lies in the interval $(45^\circ, 60^\circ)$.