For the right triangle with *a* = 18, *b* = 25, and angle∠ C=90degrees° determine the following.

Round to two decimal places.

sec *A* =?

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!

For the right triangle with *a* = 18, *b* = 25, and angle ∠C = 90°, determine the following. Round to two decimal places. sec A =?

Determine the secant of angle A in a right triangle where side a = 18, side b = 25, and angle C is 90°. This solution involves the Pythagorean theorem and trigonometric identities, suitable for students in Grades 9-11. Round your final answer to two decimal places.

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