Q Issue Q22 D Mark For Review Right triangle $$ A B C $$ is similar to right triangle $$ D E F $$, where $$ \angle C A B $$ corresponds to $$ \angle F D E $$, and $$ \angle A B C $$ and $$ \angle D E F $$ are right angles. If $$ \cos (\angle C A B)=\frac{52}{65} $$, what is the value of $$ \sec (\angle F D E) $$ ?

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UpStudy AI · Accepted Answer

To solve the problem, we need to use the properties of similar triangles and trigonometric identities.

1. **Understanding the triangles**: Since triangle $$ A B C $$ is similar to triangle $$ D E F $$, the angles of the triangles are equal. Therefore, we have:
   - $$ \angle C A B = \angle F D E $$
   - $$ \angle A B C = \angle D E F = 90^\circ $$

2. **Using the cosine value**: We are given that $$ \cos(\angle C A B) = \frac{52}{65} $$. Since $$ \angle C A B $$ corresponds to $$ \angle F D E $$, we can say:
   $$
   \cos(\angle F D E) = \cos(\angle C A B) = \frac{52}{65}
   $$

3. **Finding the secant**: The secant function is the reciprocal of the cosine function. Therefore, we can find $$ \sec(\angle F D E) $$ as follows:
   $$
   \sec(\angle F D E) = \frac{1}{\cos(\angle F D E)} = \frac{1}{\frac{52}{65}} = \frac{65}{52}
   $$

Thus, the value of $$ \sec(\angle F D E) $$ is $$ \frac{65}{52} $$.

To summarize, the final answer is:
$$
\sec(\angle F D E) = \frac{65}{52}
$$
$$ \sec(\angle F D E) = \frac{65}{52} $$

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