To find $\sec(A)$, we need to use the trigonometric definition of secant, specifically:

$\sec(A) = \frac{1}{\cos(A)}$

From the triangle, we can use the definition of cosine for angle $A$ in terms of the side lengths:

$\cos(A) = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{b}{c}$

Step 1: Use the Pythagorean Theorem to Find $c$

Since $a = 11$ and $b = 4$, we can find $c$ (the hypotenuse) using the Pythagorean theorem:

$c = \sqrt{a^2 + b^2} = \sqrt{11^2 + 4^2} = \sqrt{121 + 16} = \sqrt{137}$

So, $c = \sqrt{137}$.

Step 2: Find $\cos(A)$

Now, substitute $b$ and $c$ into the cosine formula:

$\cos(A) = \frac{b}{c} = \frac{4}{\sqrt{137}}$

Step 3: Find $\sec(A)$

Now, $\sec(A)$ is the reciprocal of $\cos(A)$:

$\sec(A) = \frac{1}{\cos(A)} = \frac{1}{\frac{4}{\sqrt{137}}} = \frac{\sqrt{137}}{4}$

Final Answer

$\sec(A) = \frac{\sqrt{137}}{4}$

Would you like further details on any step?

Related Questions:

1. How do we find $\tan(A)$ given $a$ and $b$?
2. What is the value of $\sec(B)$ in this triangle?
3. How would the solution change if $a$ and $b$ were different values?
4. Can we use trigonometric identities to verify $\sec(A)$ in a different way?
5. How do we find $\sin(A)$ in this triangle?

Tip:

For right triangles, remember that trigonometric ratios like $\sin$, $\cos$, and $\tan$ can be directly calculated from the side lengths, helping you solve for other trigonometric functions easily.