Question : $\triangle \mathrm{ABC}$ is a right-angle triangle at B and $\tan \mathrm{A}=\frac{3}{4}$, then $\sin A + \sin B + \sin C$ will be equal to:
Option 1: $2 \frac{4}{5}$
Option 2: $2 \frac{2}{5}$
Option 3: $3 \frac{1}{5}$
Option 4: $1 \frac{1}{5}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
Correct Answer: $2 \frac{2}{5}$
Solution :
In $\triangle \mathrm{ABC}$, $\tan \mathrm{A}=\frac{3}{4}$
Let the length of sides $BC$ and $BA$ be $3k$ and $4k$.
By Pythagoras theorem,
$AC^2=BC^2+BA^2$
⇒ $AC^2=3k^2+4k^2$
⇒ $AC^2=25k^2$
⇒ $AC=5k$
⇒ $\sin A = \frac{BC}{AC}$
⇒ $\sin A = \frac{3k}{5k}$
⇒ $\sin A = \frac{3}{5}$
⇒ $\sin C = \frac{BA}{AC}$
⇒ $\sin A = \frac{4k}{5k}$
⇒ $\sin A = \frac{4}{5}$
So, $\sin A+ \sin B +\sin C= \frac{3}{5}+1+\frac{4}{5}= \frac{12}{5}=2\frac{2}{5}$
Hence, the correct answer is $2\frac{2}{5}$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
Upcoming Exams
Trending Articles around SSC CGL
