MCQ
If $\sec\theta=\frac{25}{7}$ then $\sin\theta=?$
- A$\frac{7}{24}$
- B$\frac{24}{7}$
- ✓$\frac{24}{25}$
- DNone of these.
✓
Answer
Correct option: C.
$\frac{24}{25}$
Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$=\frac{\text{AC}}{\text{AB}}=\frac{25}{7}$
Let $\text{AC}=25\text{k}$ and $\text{AB}=7\text{k},$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(25\text{k})^2=(7\text{k})^2+\text{BC}^2$
$\Rightarrow\text{BC}^2=\text{625k}^2-\text{49k}^2=\text{576k}^2$
$\Rightarrow\text{BC}=\text{24k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$
$\frac{\text{BC}}{\text{AC}}=\frac{24\text{k}}{\text{25k}}=\frac{24}{25}$
Then, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$=\frac{\text{AC}}{\text{AB}}=\frac{25}{7}$
Let $\text{AC}=25\text{k}$ and $\text{AB}=7\text{k},$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(25\text{k})^2=(7\text{k})^2+\text{BC}^2$
$\Rightarrow\text{BC}^2=\text{625k}^2-\text{49k}^2=\text{576k}^2$
$\Rightarrow\text{BC}=\text{24k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$
$\frac{\text{BC}}{\text{AC}}=\frac{24\text{k}}{\text{25k}}=\frac{24}{25}$
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