MCQ
`In figure below, the value of $\sec x$ is
- ✓$\frac{13}{12}$
- B$\frac{5}{12}$
- C$\frac{12}{5}$
- D$\frac{12}{13}$
✓
Answer
Correct option: A.
$\frac{13}{12}$
$\triangle \text{ABC}$ is a right angled triangle, therefore, using Pythagoras Theorem,
$A C=\sqrt{A B^2+B C^2}$
$\Rightarrow AC=\sqrt{4^2+3^2}$
$\Rightarrow AC=\sqrt{16+9}=5 \ cm \ldots \ldots\text{(i)}$
Similarly, $\triangle \text{ACD}$ is a right angled triangle, therefore, using Pythagoras Theorem,
$AD=\sqrt{AC^2+CD^2}$
$\Rightarrow AD=\sqrt{5^2+12^2} ($ Using value of from $(i))$
$\Rightarrow AD=\sqrt{25+144}=13 \ cm$
Using values of $AC$ and $AD$, we have
$\cos \alpha=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{CD}{AD}=\frac{12}{13}$
Now, $\sec \alpha=\frac{1}{\cos \alpha}=\frac{13}{12}$
$A C=\sqrt{A B^2+B C^2}$
$\Rightarrow AC=\sqrt{4^2+3^2}$
$\Rightarrow AC=\sqrt{16+9}=5 \ cm \ldots \ldots\text{(i)}$
Similarly, $\triangle \text{ACD}$ is a right angled triangle, therefore, using Pythagoras Theorem,
$AD=\sqrt{AC^2+CD^2}$
$\Rightarrow AD=\sqrt{5^2+12^2} ($ Using value of from $(i))$
$\Rightarrow AD=\sqrt{25+144}=13 \ cm$
Using values of $AC$ and $AD$, we have
$\cos \alpha=\frac{\text { Base }}{\text { Hypotenuse }}=\frac{CD}{AD}=\frac{12}{13}$
Now, $\sec \alpha=\frac{1}{\cos \alpha}=\frac{13}{12}$
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