Question

If $sec\left(C\right)$ is $\frac{m}{2\sqrt{2}}$, then $m$ is:

• A. $1$
• B. $3$
• C. $7$
• D. none of these

Answer 1

The correct option is B

$3$

Explanation for the correct option:

Compute the required value:

From the fig, $△BDC$ is a right-angled triangle. Using Pythagoras theorem,

$$\begin{gathered} \Rightarrow D{C}^2={12}^2-4^2 \\ \Rightarrow =144-16 \\ \Rightarrow =128 \\ \Rightarrow DC=\sqrt{128}=8\sqrt{2} \end{gathered}$$

Now, $sec\left(C\right)$ $=$ hypotenuse $/$adjacent side.

$$\begin{gathered} \Rightarrow sec\left(C\right)=\frac{BC}{DC} \\ \Rightarrow sec\left(C\right)=\frac{12}{8\sqrt{2}} \\ \Rightarrow sec\left(C\right)=\frac{3}{2\sqrt{2}} \end{gathered}$$

Given: $sec\left(C\right)=\frac{m}{2\sqrt{2}}$

$$\begin{gathered} \Rightarrow \frac{m}{2\sqrt{2}}=\frac{3}{2\sqrt{2}} \\ \Rightarrow m=3 \end{gathered}$$

Hence, option $B$ is the answer.