17 of 17 Consider angles A, B, and C such that cos Acos 217", tan Btan 298°, sin C sin 7° where 0° SAS 360°,0" Suppose that A 217", B The value of A+B+Cis 210 434° 143⁰ 556° B≤ 360°,0 ≤CS 360°. 298", and CT
17 of 17 Consider angles A, B, and C such that cos Acos 217", tan Btan 298°, sin C sin 7° where 0° SAS 360°,0" Suppose that A 217", B The value of A+B+Cis 210 434° 143⁰ 556° B≤ 360°,0 ≤CS 360°. 298", and CT
Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter71: The Cartesian Coordinate System
Section: Chapter Questions
Problem 4A: If A = 3650', determine sin A, cos A, tan A, cot A, csc A, and sec A. Round each answer to 5 decimal...
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Hello. Please help me understand the attached solution to the attached question. Why use 360 with cos and 180 with tan and sin. Thanks,
![17 of 17
Consider angles A, B, and C such that cos A = cos 217", tan B = tan 298°, sin C sin 7°
where 0° SA≤ 360°,0° ≤B≤360°,0 ≤C≤ 360°.
Suppose that A 217, B #298°. and CT
The value of A+B+Cis
210°
434°
143°
556°](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa267ea74-fd2b-43ac-8dd3-20480bcefdd8%2F572ffd2a-a95e-4c1e-8cac-10b78063fa7a%2F2h5ufw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:17 of 17
Consider angles A, B, and C such that cos A = cos 217", tan B = tan 298°, sin C sin 7°
where 0° SA≤ 360°,0° ≤B≤360°,0 ≤C≤ 360°.
Suppose that A 217, B #298°. and CT
The value of A+B+Cis
210°
434°
143°
556°
![Step 2
We have,
cos A = cos 217°
cos A = cos (360° - 143°)
cos A = cos (143°)
⇒ A= 143°
tan B = tan 298°
tan B = tan (180° + 118°)
tan B = tan (118⁰)
⇒ B= 118°
sin C = sin 7°
sin C = sin (180° - 173⁰)
⇒ sin C = sin (173)
⇒ C = 173°
Now,
A + B + C = 143° + 118° + 173⁰
⇒A+B+C = 434°
Hence, the right option is B) 434°
Answer
(Answer)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa267ea74-fd2b-43ac-8dd3-20480bcefdd8%2F572ffd2a-a95e-4c1e-8cac-10b78063fa7a%2Fslazes_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Step 2
We have,
cos A = cos 217°
cos A = cos (360° - 143°)
cos A = cos (143°)
⇒ A= 143°
tan B = tan 298°
tan B = tan (180° + 118°)
tan B = tan (118⁰)
⇒ B= 118°
sin C = sin 7°
sin C = sin (180° - 173⁰)
⇒ sin C = sin (173)
⇒ C = 173°
Now,
A + B + C = 143° + 118° + 173⁰
⇒A+B+C = 434°
Hence, the right option is B) 434°
Answer
(Answer)
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Follow-up Questions
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Follow-up Question
How does "COSA = Cos360-143" turn into "COSA=Cos143"? Is it because COS360 is positive that it just cancels/zeros out?
Do you have a lesson plan regarding this? The teacher didn't teach this but its a question in my homework.
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