Question : If $\cos (A-B)=\frac{\sqrt{3}}{2}$, and $\cos (A+B)=0$, where $A$ and $B$ are positive acute angles and $A \geq B$, then the measures of $A$ and $B$ are:
Option 1: 80° and 10°
Option 2: 60° and 30°
Option 3: 70° and 20°
Option 4: 50° and 40°
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Correct Answer: 60° and 30°
Solution : Given, $\cos (A-B)=\frac{\sqrt{3}}{2}$ and $\cos (A+B)=0$ Since $A$ and $B$ are acute positive angles, $\cos (A-B)=\frac{\sqrt{3}}{2}$ ⇒ $(A-B) = \cos^{-1} \frac{\sqrt{3}}{2}$ ⇒ $(A-B) = 30°$ ------------------ (1) $\cos (A+B)=0$ ⇒ $(A+B) = \cos^{-1} 0$ ⇒ $(A+B) = 90°$ --------------------- (2) Solving (1) and (2), we get, $A = 60°$ and $B = 30°$ Hence, the correct answer is 60° and 30°.
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Question : Possible measures of the three angles of a triangle are:
Option 1: 33°, 42°, and 115°
Option 2: 40°, 70°, and 80°
Option 3: 30°, 60°, and 100°
Option 4: 50°, 60°, and 70°
Question : What will be the value of $\frac{\sin 30^{\circ} \sin 40^{\circ} \sin 50^{\circ} \sin 60^{\circ}}{\cos 30^{\circ} \cos 40^{\circ} \cos 50^{\circ} \cos 60^{\circ}}$?
Option 1: $\frac{1}{\sqrt{2}}$
Option 2: $\sqrt{3}$
Option 3: $1$
Option 4: $\frac{1}{\sqrt{3}}$
Question : If $\tan (A+B)=\sqrt{3}$ and $\tan (A-B)=\frac{1}{\sqrt{3}}; 0°<(A+B)<90°; A > B$, then the values of $A$ and $B$ are respectively:
Option 1: 45° and 15°
Option 2: 15° and 45°
Option 3: 30° and 30°
Option 4: 60° and 30°
Question : If $\theta>0$ be an acute angle, then the value of $\theta$ in degrees satisfying $\frac{\cos^2\theta-3 \cos\theta+2}{\sin^2\theta}=1$ is:
Option 1: 90°
Option 2: 30°
Option 3: 45°
Option 4: 60°
Question : If $\tan \frac{A}{2}=x$, then find $x$.
Option 1: $\frac{\sqrt{1+\cos A}}{\sqrt{1-\cos A}}$
Option 2: $\frac{\sqrt{1-\sin A}}{\sqrt{1+\cos A}}$
Option 3: $\frac{\sqrt{1-\cos A}}{\sqrt{1+\cos A}}$
Option 4: $\frac{\sqrt{\cos A-1}}{\sqrt{1+\cos A}}$
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