Trigonometric Ratios of Standard Angles Quiz - ICSE Class 9 Mathematics
Free practice quiz: Trigonometric Ratios of Standard Angles (ICSE Class 9 Mathematics). 45 multiple-choice questions with answers and explanations. Play online, no sign-up needed.
Play all 45 questions freeSample questions
If $\sin \theta = \frac{\sqrt{3}}{2}$ and $\theta$ is an acute angle, then find the value of $2\cos \theta$.
- 1
- 2
- $\frac{1}{2}$
- $\sqrt{3}$
Show answer
Answer: A. 1
Given $\sin \theta = \frac{\sqrt{3}}{2}$, for an acute angle $\theta$, we know that $\theta = 60^\circ$. Then, $\cos \theta = \cos 60^\circ = \frac{1}{2}$. Therefore, $2\cos \theta = 2 \times \frac{1}{2} = 1$.
Calculate: $\cos^2 0^\circ + \sin^2 90^\circ + 2 \tan 45^\circ$.
- 2
- 3
- 4
- 5
Show answer
Answer: C. 4
We know that $\cos 0^\circ = 1$, $\sin 90^\circ = 1$, and $\tan 45^\circ = 1$. So, $\cos^2 0^\circ = 1^2 = 1$, $\sin^2 90^\circ = 1^2 = 1$. The expression becomes $1 + 1 + 2(1) = 1 + 1 + 2 = 4$.
If $\sin x = \frac{\sqrt{3}}{2}$ and $\cos x = \frac{1}{2}$, what is $\tan x$?
- $\frac{1}{\sqrt{3}}$
- 1
- $\sqrt{3}$
- 2
Show answer
Answer: C. $\sqrt{3}$
We know that $\tan x = \frac{\sin x}{\cos x}$. Given $\sin x = \frac{\sqrt{3}}{2}$ and $\cos x = \frac{1}{2}$, $\tan x = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{2} \times \frac{2}{1} = \sqrt{3}$.
Calculate: $\sin^2 45^\circ + \cos^2 45^\circ + \tan^2 45^\circ$.
- 1
- 2
- 3
- 4
Show answer
Answer: B. 2
Using the identity $\sin^2 \theta + \cos^2 \theta = 1$, we have $\sin^2 45^\circ + \cos^2 45^\circ = 1$. We also know that $\tan 45^\circ = 1$, so $\tan^2 45^\circ = 1^2 = 1$. Therefore, the expression is $1 + 1 = 2$.
If $\tan heta = \frac{1}{\sqrt{3}}$ and $ heta$ is acute, find $\cos heta$.
- 1
- $\frac{1}{2}$
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{\sqrt{2}}$
Show answer
Answer: C. $\frac{\sqrt{3}}{2}$
If $ an heta = \frac{1}{\sqrt{3}}$ and $ heta$ is acute, then $ heta = 30^\circ$. Therefore, $\cos heta = \cos 30^\circ = \frac{\sqrt{3}}{2}$.