Question / Solve ΔΜΝΟ. Hint: This means you must find all missing sides and angles, which includes m 20 and the lengths of m and n. O n 26° m M 'N 15 cm Include all relevant calculations, and use the correct units for all measures. Be sure explanations are written in complete sentences.
Question / Solve ΔΜΝΟ. Hint: This means you must find all missing sides and angles, which includes m 20 and the lengths of m and n. O n 26° m M 'N 15 cm Include all relevant calculations, and use the correct units for all measures. Be sure explanations are written in complete sentences.
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
Related questions
Question
In the diagram, \( \triangle MNO \) is a right triangle where:
- \( \angle M \) is \( 26^\circ \),
- \( MN = 15 \) cm,
- \( \angle N \) is \( 90^\circ \),
- and you need to find the lengths of \( m \) and \( n \) and the measure of \( \angle O \).
**Include all relevant calculations, and use the correct units for all measures. Be sure explanations are written in complete sentences.**
**Solution Steps:**
1. **Determine \( \angle O \)**:
Since the sum of angles in a triangle is \( 180^\circ \) and we know two angles:
\[
\angle O = 180^\circ - \angle M - \angle N
\]
\[
\angle O = 180^\circ - 26^\circ - 90^\circ = 64^\circ
\]
2. **Find \( m \) (length of ON)**:
Using the tangent function in trigonometry:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
For \( \angle M = 26^\circ \):
\[
\tan(26^\circ) = \frac{ON}{MN} \Rightarrow \tan(26^\circ) = \frac{m}{15} \Rightarrow m = 15 \times \tan(26^\circ)
\]
Using a calculator:
\[
\tan(26^\circ) \approx 0.4877 \Rightarrow m \approx 15 \times 0.4877 \approx 7.32 \text{ cm}
\]
3. **Find \( n \) (length of OM)**:
Using the Pythagorean theorem:
\[
MN^2 + ON^2 = OM^2 \Rightarrow 15^2 + 7.32^2 = n^2
\]
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01db270a-c525-4661-be5b-2b24ecb0478b%2Ff7125b2a-3489-493d-a875-9c430c562ce7%2F3f8bhds_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 7**
**Solve \( \triangle MNO \).**
_Hint:_ This means you must find all missing sides and angles, which includes \( m \angle O \) and the lengths of \( m \) and \( n \).
In the diagram, \( \triangle MNO \) is a right triangle where:
- \( \angle M \) is \( 26^\circ \),
- \( MN = 15 \) cm,
- \( \angle N \) is \( 90^\circ \),
- and you need to find the lengths of \( m \) and \( n \) and the measure of \( \angle O \).
**Include all relevant calculations, and use the correct units for all measures. Be sure explanations are written in complete sentences.**
**Solution Steps:**
1. **Determine \( \angle O \)**:
Since the sum of angles in a triangle is \( 180^\circ \) and we know two angles:
\[
\angle O = 180^\circ - \angle M - \angle N
\]
\[
\angle O = 180^\circ - 26^\circ - 90^\circ = 64^\circ
\]
2. **Find \( m \) (length of ON)**:
Using the tangent function in trigonometry:
\[
\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
\]
For \( \angle M = 26^\circ \):
\[
\tan(26^\circ) = \frac{ON}{MN} \Rightarrow \tan(26^\circ) = \frac{m}{15} \Rightarrow m = 15 \times \tan(26^\circ)
\]
Using a calculator:
\[
\tan(26^\circ) \approx 0.4877 \Rightarrow m \approx 15 \times 0.4877 \approx 7.32 \text{ cm}
\]
3. **Find \( n \) (length of OM)**:
Using the Pythagorean theorem:
\[
MN^2 + ON^2 = OM^2 \Rightarrow 15^2 + 7.32^2 = n^2
\]
\[
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Elementary Geometry For College Students, 7e
Geometry
ISBN:
9781337614085
Author:
Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:
Cengage,
Elementary Geometry for College Students
Geometry
ISBN:
9781285195698
Author:
Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:
Cengage Learning
Elementary Geometry For College Students, 7e
Geometry
ISBN:
9781337614085
Author:
Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:
Cengage,
Elementary Geometry for College Students
Geometry
ISBN:
9781285195698
Author:
Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:
Cengage Learning