7. In the diagram, ZSEP is a right angle ZSET = 30°27'40" Find mZTEP.

Elementary Geometry For College Students, 7e
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ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
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### Geometry Problem

#### Problem Statement:
In the diagram, \(\angle SEP\) is a right angle \(\angle SET = 30^\circ 27' 40''\). Find \(m \angle TEP\).

![Geometry Diagram](Placeholder_for_image_description)

#### Solution Explanation:
To find \(m \angle TEP\), we will use the fact that the sum of angles in a triangle is always \(180^\circ\).

Given:
- \(\angle SEP = 90^\circ\) (right angle)
- \(\angle SET = 30^\circ 27' 40''\)

Steps:
1. Recall that in triangle SET, the sum of the angles is \(180^\circ\).
2. Since \(\angle SEP\) is a right angle, by definition it equals \(90^\circ\).
3. Determine \(m \angle TEP\) using the equation:
   \[
   \angle SEP + \angle SET + \angle TEP = 180^\circ
   \]

   Substituting in the known values:
   \[
   90^\circ + 30^\circ 27' 40'' + \angle TEP = 180^\circ
   \]
   \[
   120^\circ 27' 40'' + \angle TEP = 180^\circ
   \]

4. To find \(\angle TEP\), subtract \(120^\circ 27' 40''\) from \(180^\circ\):
   \[
   \angle TEP = 180^\circ - 120^\circ 27' 40''
   \]

The remaining angle \(m \angle TEP\) calculates as:
\[
\angle TEP = 180^\circ - 120^\circ 27' 40'' = 59^\circ 32' 20''
\]

Thus, \(m \angle TEP = 59^\circ 32' 20''\).

#### Diagram Description:
The provided diagram consists of three points \(S\), \(T\), and \(P\) forming a triangle with point \(E\) at the vertex of the right angle (\(90^\circ\)). Point \(E\) forms two distinct angles, one being a right angle with \(S\) and \(P\) and another \(30^\circ 27' 40''\) with \(
Transcribed Image Text:### Geometry Problem #### Problem Statement: In the diagram, \(\angle SEP\) is a right angle \(\angle SET = 30^\circ 27' 40''\). Find \(m \angle TEP\). ![Geometry Diagram](Placeholder_for_image_description) #### Solution Explanation: To find \(m \angle TEP\), we will use the fact that the sum of angles in a triangle is always \(180^\circ\). Given: - \(\angle SEP = 90^\circ\) (right angle) - \(\angle SET = 30^\circ 27' 40''\) Steps: 1. Recall that in triangle SET, the sum of the angles is \(180^\circ\). 2. Since \(\angle SEP\) is a right angle, by definition it equals \(90^\circ\). 3. Determine \(m \angle TEP\) using the equation: \[ \angle SEP + \angle SET + \angle TEP = 180^\circ \] Substituting in the known values: \[ 90^\circ + 30^\circ 27' 40'' + \angle TEP = 180^\circ \] \[ 120^\circ 27' 40'' + \angle TEP = 180^\circ \] 4. To find \(\angle TEP\), subtract \(120^\circ 27' 40''\) from \(180^\circ\): \[ \angle TEP = 180^\circ - 120^\circ 27' 40'' \] The remaining angle \(m \angle TEP\) calculates as: \[ \angle TEP = 180^\circ - 120^\circ 27' 40'' = 59^\circ 32' 20'' \] Thus, \(m \angle TEP = 59^\circ 32' 20''\). #### Diagram Description: The provided diagram consists of three points \(S\), \(T\), and \(P\) forming a triangle with point \(E\) at the vertex of the right angle (\(90^\circ\)). Point \(E\) forms two distinct angles, one being a right angle with \(S\) and \(P\) and another \(30^\circ 27' 40''\) with \(
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