Class 10 Introduction to Trigonometry Mock Test (CBSE)

Topics in Introduction to Trigonometry

• Trigonometric ratios
• Trigonometric ratios of some specific angles
• Trigonometric identities

Introduction to Trigonometry practice questions with answers

1. Q1Assertion and reason·1 mark
   Assertion (A): The value of $\sin^2 30^\circ + \cos^2 30^\circ$ is $1$. Reason (R): For any acute angle $\theta$, $\sin^2 \theta + \cos^2 \theta = 1$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). Both Assertion (A) and Reason (R) are true and Reason correctly explains Assertion.+1

   Answer: (a)
2. Q2Multiple choice·1 mark
   The value of $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$ is:

   1. (a)$\sin 60^\circ$
   2. (b)$\cos 60^\circ$
   3. (c)$\tan 60^\circ$
   4. (d)$\sin 30^\circ$
   Show answer and step marks

   1. Correct option: (a) $\sin 60^\circ$. Substituting $\tan 30^\circ = \frac{1}{\sqrt{3}}$ gives $\frac{2/\sqrt{3}}{4/3} = \frac{\sqrt{3}}{2} = \sin 60^\circ$.+1

   Answer: (a)
3. Q3Multiple choice·1 mark
   Given that $\sin \theta = \frac{a}{b}$, then $\cos \theta$ is equal to:

   1. (a)$\frac{b}{\sqrt{b^2 - a^2}}$
   2. (b)$\frac{\sqrt{b^2 - a^2}}{b}$
   3. (c)$\frac{a}{\sqrt{b^2 - a^2}}$
   4. (d)$\frac{\sqrt{b^2 - a^2}}{a}$
   Show answer and step marks

   1. Correct option: (b) $\frac{\sqrt{b^2 - a^2}}{b}$. $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b}$+1

   Answer: (b)
4. Q4Assertion and reason·1 mark
   Assertion (A): For any acute angle $\theta$, $\sin^2 \theta + \cos^2 \theta = 1$. Reason (R): The value of $\sin \theta$ can never exceed $1$ for any real angle $\theta$.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true but Reason (R) is false.
   4. (d)Assertion (A) is false but Reason (R) is true.
   Show answer and step marks

   1. Correct option: (b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A). Assertion (A) is a fundamental trigonometric identity which is true for all acute angles. Reason (R) is also a true property of the sine function based on the unit circle or right triangle ratio, but it does not explain why the identity holds.+1

   Answer: (b)
5. Q5Short answer·2 marks
   In right triangle $ABC$, $\angle B = 90^\circ$, $\angle A = 60^\circ$, $AB = 1$ cm and $AC = 2$ cm. (a) Find the length of $BC$. [1] (b) Explain why $\sin A$ is equal to $\cos C$. [1]
   Show answer and step marks

   1. (a) By Pythagoras, $BC^2 = AC^2 - AB^2 = 2^2 - 1^2 = 3$+0.5
   2. (a) $BC = \sqrt{3}$ cm+0.5
   3. (b) Angle $C = 90^\circ - 60^\circ = 30^\circ$, since the angles of a triangle add up to $180^\circ$+0.5
   4. (b) $\sin A = \frac{BC}{AC}$ and $\cos C = \frac{BC}{AC}$, so $\sin A = \cos C$ (both equal $\frac{\sqrt{3}}{2}$)+0.5

   Answer: (a) $BC = \sqrt{3}$ cm; (b) $\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} = \cos C$, because $\angle A + \angle C = 90^\circ$ and both ratios use the same sides
6. Q6Short answer·2 marks
   In right triangle $ABC$, $\angle B = 90^\circ$, $AB = 12$ cm, $BC = 5$ cm and $AC = 13$ cm. (a) Write the ratio of sides that defines $\operatorname{cosec} C$. [1] (b) Find the value of $\operatorname{cosec} C$. [1]
   Show answer and step marks

   1. (a) $\operatorname{cosec} C = \frac{1}{\sin C} = \frac{\text{hypotenuse}}{\text{side opposite } C} = \frac{AC}{AB}$+1
   2. (b) $\operatorname{cosec} C = \frac{AC}{AB} = \frac{13}{12}$+1

   Answer: (a) $\operatorname{cosec} C = \frac{AC}{AB}$; (b) $\operatorname{cosec} C = \frac{13}{12}$

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