In a right triangle, the secant (sec) of an angle is the reciprocal of the cosine (cos) of that angle. This means:

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

To find $\cos A$, we use the formula:

$\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}$

Here, $a = 18$ (adjacent to angle $A$), $b = 25$ (opposite to angle $A$), and $\angle C = 90^\circ$, which means we need to find the hypotenuse $c$.

Using the Pythagorean theorem to find the hypotenuse $c$:

$c = \sqrt{a^2 + b^2} = \sqrt{18^2 + 25^2} = \sqrt{324 + 625} = \sqrt{949} \approx 30.80$

Now, we can find $\cos A$:

$\cos A = \frac{a}{c} = \frac{18}{30.80} \approx 0.5844$

Finally, the secant of angle $A$ is:

$\sec A = \frac{1}{\cos A} = \frac{1}{0.5844} \approx 1.71$

So, $\sec A \approx 1.71$.

Would you like any further details?

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Here are 5 related questions:

1. How do you calculate sine for angle $A$?
2. What is the value of $\tan A$ for this triangle?
3. Can you find $\sec B$ using similar steps?
4. What is the formula for finding the hypotenuse in a right triangle?
5. How does secant relate to other trigonometric functions?

Tip: Always ensure to use the Pythagorean theorem when missing the hypotenuse in right triangles!