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: $1 \frac{1}{5}$
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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 $2\frac{2}{5}$.
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