Here are 25 multiple-choice questions on Trigonometry, tailored for JKSSB and similar competitive exams, with answers and explanations:
Section B: Mathematics – Trigonometry
Q1. What is the value of $\sin(30^\circ)$?
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) $1$
(d) $0$
Answer: (a)
Explanation: The sine of $30^\circ$ is a standard trigonometric value, representing the ratio of the side opposite to the hypotenuse in a right-angled triangle with a $30^\circ$ angle.
Q2. If $\tan(\theta) = \frac{3}{4}$, what is the value of $\sin(\theta)$?
(a) $\frac{3}{5}$
(b) $\frac{4}{5}$
(c) $\frac{5}{3}$
(d) $\frac{5}{4}$
Answer: (a)
Explanation: If $\tan(\theta) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{3}{4}$. Using Pythagoras theorem, Hypotenuse = $\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5$. So, $\sin(\theta) = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{3}{5}$.
Q3. What is the value of $\cos(60^\circ)$?
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) $1$
(d) $0$
Answer: (a)
Explanation: The cosine of $60^\circ$ is a standard trigonometric value.
Q4. The angle of elevation of the top of a tower from a point on the ground, which is $30$ m away from the foot of the tower, is $30^\circ$. What is the height of the tower?
(a) $10\sqrt{3}$ m
(b) $30\sqrt{3}$ m
(c) $10$ m
(d) $30$ m
Answer: (a)
Explanation: Let ‘h’ be the height of the tower. We have $\tan(30^\circ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{30}$. Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we have $\frac{h}{30} = \frac{1}{\sqrt{3}}$. So, $h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3}$ m.
Q5. Which of the following is equivalent to $\cot(\theta)$?
(a) $\frac{\sin(\theta)}{\cos(\theta)}$
(b) $\frac{1}{\sec(\theta)}$
(c) $\frac{\cos(\theta)}{\sin(\theta)}$
(d) $\sin(\theta) \cdot \cos(\theta)$
Answer: (c)
Explanation: The cotangent function is defined as the reciprocal of the tangent function, and $\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$. Therefore, $\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)}$.
Q6. If $\sin(\theta) = \cos(\theta)$, then what is the value of $\theta$ in the range $0^\circ \le \theta \le 90^\circ$?
(a) $30^\circ$
(b) $45^\circ$
(c) $60^\circ$
(d) $90^\circ$
Answer: (b)
Explanation: $\sin(\theta) = \cos(\theta)$ implies $\frac{\sin(\theta)}{\cos(\theta)} = 1$, which means $\tan(\theta) = 1$. The angle for which $\tan(\theta) = 1$ in the given range is $45^\circ$.
Q7. What is the value of $\sec^2(\theta) – \tan^2(\theta)$?
(a) $0$
(b) $1$
(c) $-1$
(d) $\sin^2(\theta)$
Answer: (b)
Explanation: This is one of the fundamental trigonometric identities: $1 + \tan^2(\theta) = \sec^2(\theta)$. Rearranging it gives $\sec^2(\theta) – \tan^2(\theta) = 1$.
Q8. The maximum value of $\sin(\theta)$ is:
(a) $0$
(b) $1$
(c) $\frac{1}{2}$
(d) $\sqrt{3}$
Answer: (b)
Explanation: The range of the sine function is $[-1, 1]$. Therefore, its maximum value is $1$. This occurs at $\theta = 90^\circ$.
Q9. If $\sin(\theta) = \frac{1}{\sqrt{2}}$, what is the value of $\cot(\theta)$?
(a) $\frac{1}{\sqrt{2}}$
(b) $\sqrt{2}$
(c) $1$
(d) $0$
Answer: (c)
Explanation: If $\sin(\theta) = \frac{1}{\sqrt{2}}$, then $\theta = 45^\circ$. For $\theta = 45^\circ$, $\cot(45^\circ) = 1$. Alternatively, if $\sin(\theta) = \frac{1}{\sqrt{2}}$, then $\cos(\theta) = \sqrt{1 – \sin^2(\theta)} = \sqrt{1 – (\frac{1}{\sqrt{2}})^2} = \sqrt{1 – \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$. So, $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{1/\sqrt{2}}{1/\sqrt{2}} = 1$.
Q10. What is $\cos(0^\circ)$?
(a) $0$
(b) $1$
(c) $\frac{1}{2}$
(d) Undefined
Answer: (b)
Explanation: The cosine of $0^\circ$ is a standard trigonometric value.
Q11.. In a right-angled triangle, if one acute angle is $50^\circ$, what is the measure of the other acute angle?
(a) $30^\circ$
(b) $40^\circ$
(c) $60^\circ$
(d) $50^\circ$
Answer: (b)
Explanation: The sum of angles in a triangle is $180^\circ$. In a right-angled triangle, one angle is $90^\circ$. So, the sum of the two acute angles is $180^\circ – 90^\circ = 90^\circ$. If one acute angle is $50^\circ$, the other is $90^\circ – 50^\circ = 40^\circ$.
Q12. What is the value of $\sin(90^\circ – \theta)$?
(a) $\sin(\theta)$
(b) $\cos(\theta)$
(c) $\tan(\theta)$
(d) $\cot(\theta)$
Answer: (b)
Explanation: This is a complementary angle identity: $\sin(90^\circ – \theta) = \cos(\theta)$.
Q13. If $\tan(A) = 1$, then $A$ is:
(a) $0^\circ$
(b) $30^\circ$
(c) $45^\circ$
(d) $60^\circ$
Answer: (c)
Explanation: The angle whose tangent is $1$ is $45^\circ$.
Q14. The angle of depression from the top of a $20$ m high building to a point on the ground is $45^\circ$. How far is the point from the base of the building?
(a) $10$ m
(b) $20$ m
(c) $20\sqrt{2}$ m
(d) $10\sqrt{3}$ m
Answer: (b)
Explanation: The angle of depression from the top of the building to the point on the ground is equal to the angle of elevation from the point on the ground to the top of the building (alternate interior angles). Let ‘x’ be the distance. $\tan(45^\circ) = \frac{\text{height}}{\text{distance}} = \frac{20}{x}$. Since $\tan(45^\circ) = 1$, we have $1 = \frac{20}{x}$, so $x = 20$ m.
Q15. Which of the following is equivalent to $\csc(\theta)$?
(a) $\frac{1}{\sin(\theta)}$
(b) $\frac{1}{\cos(\theta)}$
(c) $\frac{1}{\tan(\theta)}$
(d) $\frac{1}{\sec(\theta)}$
Answer: (a)
Explanation: The cosecant function is defined as the reciprocal of the sine function.
Q16. What is the value of $\cos(90^\circ)$?
(a) $0$
(b) $1$
(c) $-1$
(d) Undefined
Answer: (a)
Explanation: The cosine of $90^\circ$ is a standard trigonometric value.
Q17. If $\sec(\theta) = 2$, what is the value of $\cos(\theta)$?
(a) $2$
(b) $\frac{1}{2}$
(c) $1$
(d) $0$
Answer: (b)
Explanation: Since $\sec(\theta) = \frac{1}{\cos(\theta)}$, if $\sec(\theta) = 2$, then $\cos(\theta) = \frac{1}{2}$.
Q18. In a right-angled triangle, if the hypotenuse is $10$ units and an angle is $30^\circ$, what is the length of the side adjacent to the $30^\circ$ angle?
(a) $5$ units
(b) $5\sqrt{3}$ units
(c) $10\sqrt{3}$ units
(d) $10$ units
Answer: (b)
Explanation: Let ‘x’ be the adjacent side. We have $\cos(30^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{10}$. Since $\cos(30^\circ) = \frac{\sqrt{3}}{2}$, we have $\frac{x}{10} = \frac{\sqrt{3}}{2}$. So, $x = \frac{10\sqrt{3}}{2} = 5\sqrt{3}$ units.
Q19. Which of the following identity is correct?
(a) $\sin^2(\theta) – \cos^2(\theta) = 1$
(b) $\sin^2(\theta) + \cos^2(\theta) = 1$
(c) $\tan^2(\theta) – \sec^2(\theta) = 1$
(d) $\cot^2(\theta) – \csc^2(\theta) = 1$
Answer: (b)
Explanation: This is the fundamental trigonometric identity based on the Pythagorean theorem.
Q20. What is the value of $\tan(0^\circ)$?
(a) $0$
(b) $1$
(c) Undefined
(d) $\sqrt{3}$
Answer: (a)
Explanation: $\tan(0^\circ) = \frac{\sin(0^\circ)}{\cos(0^\circ)} = \frac{0}{1} = 0$.
Q21. If $\sin(A) = \frac{5}{13}$, then $\cos(A)$ is:
(a) $\frac{12}{13}$
(b) $\frac{13}{5}$
(c) $\frac{5}{12}$
(d) $\frac{13}{12}$
Answer: (a)
Explanation: Using the identity $\sin^2(A) + \cos^2(A) = 1$, we have $\cos^2(A) = 1 – \sin^2(A) = 1 – (\frac{5}{13})^2 = 1 – \frac{25}{169} = \frac{169-25}{169} = \frac{144}{169}$. So, $\cos(A) = \sqrt{\frac{144}{169}} = \frac{12}{13}$ (assuming A is in the first quadrant where cosine is positive).
Q22. A ladder $15$ m long just reaches the top of a vertical wall. If the ladder makes an angle of $60^\circ$ with the wall, what is the height of the wall?
(a) $15\sqrt{3}$ m
(b) $\frac{15}{2}$ m
(c) $15$ m
(d) $15\sqrt{2}$ m
Answer: (b)
Explanation: Let ‘h’ be the height of the wall. The angle the ladder makes with the wall is $60^\circ$. This means the angle the ladder makes with the ground is $90^\circ – 60^\circ = 30^\circ$.
Using the angle with the wall: $\cos(60^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{h}{15}$. Since $\cos(60^\circ) = \frac{1}{2}$, we have $\frac{h}{15} = \frac{1}{2}$. So, $h = \frac{15}{2}$ m.
Alternatively, using the angle with the ground ($30^\circ$): $\sin(30^\circ) = \frac{h}{15}$. $\frac{1}{2} = \frac{h}{15}$, so $h = \frac{15}{2}$ m.
Q23. What is the value of $\cos^2(45^\circ) – \sin^2(30^\circ)$?
(a) $1$
(b) $\frac{1}{4}$
(c) $\frac{1}{2}$
(d) $\frac{3}{4}$
Answer: (b)
Explanation: $\cos(45^\circ) = \frac{1}{\sqrt{2}}$, so $\cos^2(45^\circ) = (\frac{1}{\sqrt{2}})^2 = \frac{1}{2}$.
$\sin(30^\circ) = \frac{1}{2}$, so $\sin^2(30^\circ) = (\frac{1}{2})^2 = \frac{1}{4}$.
Therefore, $\cos^2(45^\circ) – \sin^2(30^\circ) = \frac{1}{2} – \frac{1}{4} = \frac{2-1}{4} = \frac{1}{4}$.
Q24. If $\cot(A) = \frac{8}{15}$, then $\tan(A)$ is:
(a) $\frac{15}{8}$
(b) $\frac{8}{17}$
(c) $\frac{15}{17}$
(d) $\text{Cannot be determined}$
Answer: (a)
Explanation: $\tan(A)$ is the reciprocal of $\cot(A)$. So, $\tan(A) = \frac{1}{\cot(A)} = \frac{1}{8/15} = \frac{15}{8}$.
Q25. The value of $\frac{\sin^2(60^\circ) + \cos^2(60^\circ)}{\tan^2(45^\circ)}$ is:
(a) $1$
(b) $2$
(c) $\frac{1}{2}$
(d) $0$
Answer: (a)
Explanation: We know that $\sin^2(\theta) + \cos^2(\theta) = 1$ for any angle $\theta$. So, $\sin^2(60^\circ) + \cos^2(60^\circ) = 1$.
Also, $\tan(45^\circ) = 1$, so $\tan^2(45^\circ) = 1^2 = 1$.
Therefore, the expression becomes $\frac{1}{1} = 1$.
