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

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section: Chapter Questions
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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°
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)
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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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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