1. A ladder leans against a vertical wall at an angle of $60^\circ$ with the ground. If the foot of the ladder is $5\text{ m}$ away from the wall, what is the length of the ladder?

$5\text{ m}$

$10\text{ m}$

$5\sqrt{3}\text{ m}$

$10\sqrt{3}\text{ m}$

Solution:
Let the length of the ladder be $L$.
The distance from the wall (base) is $5\text{ m}$.
Using the cosine ratio:
$$\cos 60^\circ = \frac{\text{Base}}{\text{Hypotenuse}}$$
$$\frac{1}{2} = \frac{5}{L}$$
$$L = 5 \times 2 = 10\text{ m}$$.

2. What is the value of the expression $\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \cdot ... \cdot \cos 100^\circ$?

$1$

$-1$

$\frac{1}{\sqrt{2}}$

$0$

Solution:
The series includes the term $\cos 90^\circ$.
We know that $\cos 90^\circ = 0$.
Since all terms are multiplied together, the presence of a zero makes the entire product zero.
$$\text{Product} = (\cos 1^\circ \cdot ... \cdot \cos 89^\circ) \times 0 \times (\cos 91^\circ \cdot ... \cdot \cos 100^\circ) = 0$$.

3. If $\tan A = \cot B$, where $A$ and $B$ are acute angles, what is the value of $(A + B)$?

$90^\circ$

$45^\circ$

$180^\circ$

$0^\circ$

Solution:
We know the complementary identity: $\cot B = \tan(90^\circ - B)$.
Given $\tan A = \cot B$, we can write:
$$\tan A = \tan(90^\circ - B)$$
Comparing the angles:
$$A = 90^\circ - B \implies A + B = 90^\circ$$.

4. Find the value of $\frac{\sin 18^\circ}{\cos 72^\circ} + \frac{\tan 25^\circ}{\cot 65^\circ}$.

$0$

$1$

$2$

$4$

Solution:
Using complementary angles:
$\cos 72^\circ = \cos(90^\circ - 18^\circ) = \sin 18^\circ$.
$\cot 65^\circ = \cot(90^\circ - 25^\circ) = \tan 25^\circ$.
Substituting these in the expression:
$$\frac{\sin 18^\circ}{\sin 18^\circ} + \frac{\tan 25^\circ}{\tan 25^\circ} = 1 + 1 = 2$$.

5. If $\sec \theta + \tan \theta = 3$, find the value of $\sec \theta - \tan \theta$.

$3$

$1/3$

$-3$

$0$

Solution:
We know the identity: $\sec^2 \theta - \tan^2 \theta = 1$.
This can be factored as $(\sec \theta + \tan \theta)(\sec \theta - \tan \theta) = 1$.
Given $(\sec \theta + \tan \theta) = 3$, we have:
$$3 \times (\sec \theta - \tan \theta) = 1$$
$$\sec \theta - \tan \theta = \frac{1}{3}$$.

6. The shadow of a vertical pole is equal to its height. What is the angle of elevation of the sun?

$30^\circ$

$60^\circ$

$90^\circ$

$45^\circ$

Solution:
Let the height of the pole be $h$ and the length of the shadow be $s$.
Given $h = s$.
If $\theta$ is the angle of elevation:
$$\tan \theta = \frac{\text{Height}}{\text{Shadow}} = \frac{h}{s}$$
Since $h = s$, $\tan \theta = 1$.
We know $\tan 45^\circ = 1$, so $\theta = 45^\circ$.

7. Simplify the expression: $(\sin^2 \theta + \cos^2 \theta) \cdot (\sec^2 \theta - \tan^2 \theta)$.

$1$

$0$

$2$

$-1$

Solution:
We use basic trigonometric identities:
1. $\sin^2 \theta + \cos^2 \theta = 1$
2. $\sec^2 \theta - \tan^2 \theta = 1$
Therefore:
$$1 \times 1 = 1$$.

8. Convert $60^\circ$ into radians.

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\frac{\pi}{3}$

$\frac{\pi}{6}$

Solution:
To convert degrees to radians, multiply by $\frac{\pi}{180}$.
$$\text{Radians} = 60 \times \frac{\pi}{180}$$
$$\text{Radians} = \frac{\pi}{3}$$.

9. If $x = 5 \sin \theta$ and $y = 5 \cos \theta$, what is the value of $x^2 + y^2$?

$5$

$25$

$1$

$10$

Solution:
Square both equations and add them:
$x^2 = 25 \sin^2 \theta$
$y^2 = 25 \cos^2 \theta$
$$x^2 + y^2 = 25 \sin^2 \theta + 25 \cos^2 \theta$$
$$x^2 + y^2 = 25 (\sin^2 \theta + \cos^2 \theta)$$
Since $\sin^2 \theta + \cos^2 \theta = 1$:
$$x^2 + y^2 = 25(1) = 25$$.

10. Find the maximum value of $2 + \sin \theta$.

$3$

$2$

$1$

$0$

Solution:
The range of $\sin \theta$ is $[-1, 1]$.
The maximum value of $\sin \theta$ is $1$.
Therefore, the maximum value of $2 + \sin \theta$ is:
$$2 + 1 = 3$$.

Competitive Exams : Quantitative Aptitude - Trig (Set 3) | Concept of Maths

Competitive Exams : Quantitative Aptitude - Trig (Set 3) | Concept of Maths

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