Question : If $\tan A=\frac{2}{3}$, then find $\sin A$.
Option 1: $\frac{1}{3}$
Option 2: $\frac{2}{\sqrt{13}}$
Option 3: $\frac{2}{3}$
Option 4: $\frac{3}{\sqrt{13}}$
Latest: SSC CGL Tier 1 Result 2024 Out | SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL Tier 1 Scorecard 2024 Released | SSC CGL complete guide
Suggested: Month-wise Current Affairs | Upcoming Government Exams
Correct Answer: $\frac{2}{\sqrt{13}}$
Solution : Given, $\tan A=\frac{2}{3}$ Let perpendicular and base be 2 and 3 units. Now, the hypotenuse $= \sqrt{2^2+3^2} = \sqrt{13}$ units So, $\sin A = \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{2}{\sqrt{13}}$ Hence, the correct answer is $\frac{2}{\sqrt{13}}$.
Candidates can download this ebook to know all about SSC CGL.
Result | Eligibility | Application | Selection Process | Preparation Tips | Admit Card | Answer Key
Question : If $2 \cot \theta = 3$, find the value of $\frac{\sqrt{13} \sin \theta – 3 \tan \theta}{3 \tan \theta + \sqrt{13} \cos \theta}$
Question : If $\tan \frac{A}{2}=x$, then find $x$.
Question : Using $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}$, find the value of $\tan 15°$.
Question : If $\frac{\sin ^2 \theta}{\cos ^2 \theta-3 \cos \theta+2}=1, \theta$ lies in the first quadrant, then the value of $\frac{\tan ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}}{\tan \theta+\sin \theta}$ is:
Question : If $\tan \frac{\pi}{6}+\sec\frac{\pi}{6}=x$, then find $x$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile