MCQ
If $\sin\text{A} = \frac{12}{13}$ then $\tan\text{A} = $
- A$\frac{13}{5}$
- ✓$\frac{12}{5}$
- C$\frac{13}{2}$
- D$\frac{5}{12}$
✓
Answer
Correct option: B.
$\frac{12}{5}$
Let $BC = 12\ cm$ and $AB = 13\ cm$
$\therefore\text{AC} = \sqrt{(13\text{k})^{2}+(12\text{k})^{2}} $
$= \sqrt{169\text{k}}^{2} - 144\text{k}^{2}$
$\Rightarrow\text{AC} = \sqrt{25\text{k}^{2}} = 5\text{k}$
$\therefore\tan\text{A} = \frac{12\text{k}}{5\text{k}} = \frac{12}{5}$
$\therefore\text{AC} = \sqrt{(13\text{k})^{2}+(12\text{k})^{2}} $
$= \sqrt{169\text{k}}^{2} - 144\text{k}^{2}$
$\Rightarrow\text{AC} = \sqrt{25\text{k}^{2}} = 5\text{k}$
$\therefore\tan\text{A} = \frac{12\text{k}}{5\text{k}} = \frac{12}{5}$
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