Question : In the given figure, O is the centre of the circle, OQ is perpendicular to RS, and $\angle SRT=30^{\circ}$. If $RS=10\sqrt{2}$ units, what is the value of PR2?
Option 1: $200\left ( 1+\sqrt{3} \right )$ sq. units
Option 2: $300\left ( 2+\sqrt{3} \right )$ sq. units
Option 3: $200\left ( 2+\sqrt{3} \right )$ sq. units
Option 4: $100\left ( 3+2\sqrt{3} \right )$ sq. units
Correct Answer: $200\left ( 2+\sqrt{3} \right )$ sq. units
Solution : Here RQ = QS = $5\sqrt{2}$ $\angle$SRT = 30° In $\triangle$ROQ, $\angle$ORQ = 90° – 30° = 60° ⇒ OR $=\frac{5\sqrt{2}}{\cos 60º} = 10\sqrt{2} = OP$ OQ $=5\sqrt{2} \times \tan 60º = 5\sqrt{6}$ PQ = OP + OQ = $10\sqrt{2} + 5\sqrt{6}$ In $\triangle$PQR, PR2 = RQ2 + PQ2 = $(5\sqrt{2})^2 + (10\sqrt{2} + 5\sqrt{6})^2$ = $50 + 200 + 150 + 100\sqrt{12}$ = $400 + 200\sqrt{3} = 200(2 + \sqrt{3})$ sq. units Hence, the correct answer is $200\left ( 2+\sqrt{3} \right )$ sq. units.
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Question : Three circles of equal radius '$a$' cm touch each other. The area of the shaded region is:
Option 1: $\left ( \frac{\sqrt{3}+\pi}{2} \right )a^{2}$ sq. cm
Option 2: $\left ( \frac{6\sqrt{3}-\pi }{2} \right )a^{2}$ sq. cm
Option 3: $\left ( \sqrt{3}-\pi \right )a^{2}$ sq. cm
Option 4: $\left ( \frac{2\sqrt{3}- \pi}{2} \right )a^{2}$ sq. cm
Question : If $k\left(\tan 45^{\circ} \sin 60^{\circ}\right)=\cos 60^{\circ} \cot 30^{\circ}$, then the value of k is:
Option 1: $1$
Option 2: $2$
Option 3: $\frac{1}{\sqrt{3}}$
Option 4: $\sqrt{3}$
Question : Find the value of $\frac{\sin ^2 39^{\circ}+\sin ^2\left(90^{\circ}–39^{\circ}\right)}{\cos ^2 35^{\circ}+\cos ^2\left(90^{\circ}–35^{\circ}\right)}+3 \tan 25^{\circ} \tan 75^{\circ}$:
Option 1: 2
Option 2: 4
Option 3: 3
Option 4: 1
Question : If $3+\cos ^2 \theta=3\left(\cot ^2 \theta+\sin ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, then what is the value of $(\cos \theta+2 \sin \theta)$ ?
Option 1: $\frac{2 \sqrt{3}+1}{2}$
Option 2: $3 \sqrt{2}$
Option 3: $\frac{3 \sqrt{3}+1}{2}$
Option 4: $\frac{\sqrt{3}+2}{2}$
Question : The value of $\tan ^2 48^{\circ}-\operatorname{cosec}^2 42^{\circ}+\operatorname{cosec}\left(67^{\circ}+\theta\right)-\sec \left(23^{\circ}-\theta\right)$ is:
Option 1: $-1$
Option 2: $0$
Option 3: $1$
Option 4: $-2$
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