Trigonometric Ratios of Some Specific Angles Questions in English

(C) Given that $A = 30^{\circ}$.
We need to find the value of $\cos 2A$.
Substitute the value of $A$ into the expression:
$\cos 2A = \cos(2 \times 30^{\circ})$
$\cos 2A = \cos 60^{\circ}$
Since $\cos 60^{\circ} = \frac{1}{2}$,the value is $\frac{1}{2}$.

(B) We know the trigonometric identity for complementary angles: $\tan \theta = \cot(90^{\circ} - \theta)$.
Substituting $\theta = 15^{\circ}$ into the identity,we get:
$\tan 15^{\circ} = \cot(90^{\circ} - 15^{\circ})$.
$\tan 15^{\circ} = \cot 75^{\circ}$.
Therefore,the value of $\tan 15^{\circ}$ is equal to $\cot 75^{\circ}$.

(A) We know the trigonometric values: $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,$\tan 45^{\circ} = 1$,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$,and $\cot 90^{\circ} = 0$.
Substituting these values into the expression:
$\sin^{2} 60^{\circ} - \tan 45^{\circ} + \cos^{2} 30^{\circ} - \cot 90^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^{2} - 1 + \left(\frac{\sqrt{3}}{2}\right)^{2} - 0$
$= \frac{3}{4} - 1 + \frac{3}{4} - 0$
$= \frac{3}{4} + \frac{3}{4} - 1$
$= \frac{6}{4} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.

(B) Given expression: $8 \sin^{2} 45^{\circ} - 2 \tan^{2} 60^{\circ} + 3 \cot^{2} 30^{\circ} - 2 \cos^{2} 45^{\circ}$
Substitute the trigonometric values:
$\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,$\tan 60^{\circ} = \sqrt{3}$,$\cot 30^{\circ} = \sqrt{3}$,$\cos 45^{\circ} = \frac{1}{\sqrt{2}}$
$= 8 \left( \frac{1}{\sqrt{2}} \right)^{2} - 2 (\sqrt{3})^{2} + 3 (\sqrt{3})^{2} - 2 \left( \frac{1}{\sqrt{2}} \right)^{2}$
$= 8 \left( \frac{1}{2} \right) - 2 (3) + 3 (3) - 2 \left( \frac{1}{2} \right)$
$= 4 - 6 + 9 - 1$
$= 6$

(C) Given that $\tan \theta = 1$.
Since $\tan 45^{\circ} = 1$,we have $\theta = 45^{\circ}$.
Now,substitute $\theta = 45^{\circ}$ into the expression $\sin \theta \cdot \cos \theta$:
$\sin 45^{\circ} \cdot \cos 45^{\circ} = \left( \frac{1}{\sqrt{2}} \right) \cdot \left( \frac{1}{\sqrt{2}} \right) = \frac{1}{2}$.

(A) Given expression: $\sin^{2} 30^{\circ} - \tan 45^{\circ} + \cos^{2} 60^{\circ} - \cot 90^{\circ}$
Substitute the trigonometric values:
$\sin 30^{\circ} = \frac{1}{2}$,$\tan 45^{\circ} = 1$,$\cos 60^{\circ} = \frac{1}{2}$,and $\cot 90^{\circ} = 0$
Substituting these values into the expression:
$= (\frac{1}{2})^{2} - 1 + (\frac{1}{2})^{2} - 0$
$= \frac{1}{4} - 1 + \frac{1}{4}$
$= \frac{2}{4} - 1$
$= \frac{1}{2} - 1$
$= -\frac{1}{2}$

(A) Given that $A$ and $B$ are acute angles.
Since $\tan A = 1,$ we know that $\tan 45^{\circ} = 1,$ therefore $A = 45^{\circ}.$
Since $\sin B = \frac{1}{\sqrt{2}},$ we know that $\sin 45^{\circ} = \frac{1}{\sqrt{2}},$ therefore $B = 45^{\circ}.$
Now,we need to find the value of $\cos (A + B).$
Substituting the values of $A$ and $B,$ we get $\cos (45^{\circ} + 45^{\circ}) = \cos 90^{\circ}.$
Since $\cos 90^{\circ} = 0,$ the final answer is $0$.

(B) In a right-angled triangle,the side opposite to the $30^{\circ}$ angle is given by the formula: $\text{Side opposite to } 30^{\circ} = \text{Hypotenuse} \times \sin(30^{\circ})$.
Given that the hypotenuse is $12$ and $\sin(30^{\circ}) = \frac{1}{2}$.
Therefore,the measure of the side opposite to the angle is $12 \times \frac{1}{2} = 6$.

(B) Consider a right-angled triangle $ABC$ where $\angle B = 90^{\circ}$ and $\angle C = 60^{\circ}$.
In this triangle,the side opposite to $\angle C$ is $AB$ and the hypotenuse is $AC$.
Using the trigonometric ratio $\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}$,we have:
$\sin 60^{\circ} = \frac{AB}{AC}$
Since $\sin 60^{\circ} = \frac{\sqrt{3}}{2}$,we get:
$\frac{\sqrt{3}}{2} = \frac{AB}{AC}$
Therefore,$AB = \frac{\sqrt{3}}{2} AC$.
Thus,the measure of the side opposite to the angle of $60^{\circ}$ is $\frac{\sqrt{3}}{2}$ times the measure of the hypotenuse.

(C) In $\Delta ABC$,$\angle B = 90^{\circ}$.
Given $AC = 16$ (hypotenuse) and $BC = 8$ (side adjacent to $\angle C$).
Using the trigonometric ratio for cosine:
$\cos C = \frac{\text{Adjacent side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{8}{16} = \frac{1}{2}$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $m \angle C = 60^{\circ}$.

(D) Let the right triangle be $\triangle ABC$,where $\angle B = 90^{\circ}$ and $\angle C = 45^{\circ}$.
In $\triangle ABC$,the side opposite to $\angle C$ is $AB$ and the hypotenuse is $AC$.
Using the trigonometric ratio $\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}$:
$\sin 45^{\circ} = \frac{AB}{AC}$
Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,we have:
$\frac{1}{\sqrt{2}} = \frac{AB}{AC}$
Therefore,$AB = \frac{1}{\sqrt{2}} AC$.
Thus,the measure of the side opposite to the angle of $45^{\circ}$ is $\frac{1}{\sqrt{2}}$ times the measure of the hypotenuse.

(D) In the right-angled $\Delta ABC$,we are given the hypotenuse $AC = 20$ and the angle $m\angle C = 30^\circ$.
We need to find the adjacent side $BC$.
Using the trigonometric ratio $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}}$,we have:
$\cos 30^\circ = \frac{BC}{AC}$
Substituting the values:
$\frac{\sqrt{3}}{2} = \frac{BC}{20}$
$BC = 20 \times \frac{\sqrt{3}}{2}$
$BC = 10\sqrt{3}$
Using $\sqrt{3} \approx 1.73$:
$BC = 10 \times 1.73 = 17.3$

(A) In right-angled $\Delta ABC$,we are given the hypotenuse $AC = 30 \text{ cm}$ and $\angle C = 30^\circ$.
We need to find the side $AB$,which is opposite to $\angle C$.
Using the trigonometric ratio $\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}$:
$\sin 30^\circ = \frac{AB}{AC}$
Since $\sin 30^\circ = \frac{1}{2}$,we have:
$\frac{1}{2} = \frac{AB}{30}$
$AB = \frac{30}{2} = 15 \text{ cm}$.