Question : In right angled triangle ABC, if $\angle \mathrm{A}=30^{\circ}$, find the value of $3 \sin \mathrm{A}-4 \sin ^3 \mathrm{~A}$.
Option 1: 2
Option 2: –1
Option 3: –2
Option 4: 1
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Correct Answer: 1
Solution : Given that $\angle A = 30^{\circ}$ We know that, $\sin A = \sin 30°= \frac{1}{2}$ Now, Substitute $\sin A = \frac{1}{2}$ into the expression $3 \sin A - 4 \sin^3 A$: $3 \sin A - 4 \sin^3 A = 3 \times \frac{1}{2} - 4 \times \left(\frac{1}{2}\right)^3 = \frac{3}{2} - \frac{4}{8} = \frac{3}{2} - \frac{1}{2} = 1$ Hence, the correct answer is 1.
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Question : Let ABC be a right-angled triangle where $\angle \mathrm{A}=90^{\circ}$ and $\angle \mathrm{C}=45^{\circ}$. Find the value of $\sec \mathrm{C}+\sin \mathrm{C} \sec \mathrm{C}$.
Option 1: $1$
Option 2: $1-\sqrt{2}$
Option 3: $1+\sqrt{2}$
Option 4: $\sqrt{2}-1$
Question : In a right-angled triangle $\mathrm{XYZ}$, if $\mathrm{X}=60^{\circ}$ and $\mathrm{Y}=30^{\circ}$, then find the value of $\sin (\mathrm{X}-\mathrm{Y})$.
Option 1: $\frac{3}{5}$
Option 2: $\frac{1}{2}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{2}{3}$
Question : In $\triangle \mathrm{ABC}, \angle \mathrm{A}=5 \mathrm{x}-60^{\circ}, \angle \mathrm{B}=2 \mathrm{x}+40^{\circ}, \angle \mathrm{C}=3 \mathrm{x}-80^{\circ}$. Find $\angle \mathrm{A}$.
Option 1: $75^{\circ}$
Option 2: $90^{\circ}$
Option 3: $80^{\circ}$
Option 4: $60^{\circ}$
Question : In a $\triangle ABC$, if $2\angle A=3\angle B=6\angle C$, then the value of $\angle B$ is:
Option 1: $60^{\circ}$
Option 2: $30^{\circ}$
Option 3: $45^{\circ}$
Option 4: $90^{\circ}$
Question : In $\triangle \mathrm{PQR}, \angle \mathrm{P}=46^{\circ}$ and $\angle \mathrm{R}=64^{\circ}$.If $\triangle \mathrm{PQR}$ is similar to $\triangle \mathrm{ABC}$ and in correspondence, then what is the value of $\angle \mathrm{B}$?
Option 1: 70°
Option 2: 90°
Option 3: 100°
Option 4: 110°
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