Question : $\triangle \mathrm{ABC}$ is a right-angle triangle at $\mathrm{B}$. If $\tan \mathrm{A}=\frac{5}{12}$, then $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}$ will be equal to:
Option 1: $1 \frac{5}{13}$
Option 2: $2 \frac{4}{13}$
Option 3: $3 \frac{1}{13}$
Option 4: $2 \frac{1}{13}$
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Correct Answer: $2 \frac{4}{13}$
Solution : Given, $\tan\mathrm{A}=\frac{5}{12}$ Using Pythagoras theorem, $AC^2 = AB^2+BC^2$ ⇒ $AC^2 = 5^2 + 12^2$ ⇒ $AC^2 = 25 + 144$ ⇒ $AC^2 = 169$ ⇒ $AC = 13$ units ⇒ $\sin \mathrm{A} = \frac{5}{13},$ $\sin \mathrm{B} = 1$ and $\sin\mathrm{C} = \frac{12}{13}$ So, $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C}$ = $\frac{5}{13}+1+\frac{12}{13}$ = $\frac{12+13+5}{13}=\frac{30}{13}=2\frac{4}{13}$ Hence, the correct answer is $2\frac{4}{13}$.
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