Why is Angle C Assumed to be Acute and What is the Value of Sin C in ∆ABC?

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Discussion Overview

The discussion revolves around the assumptions regarding angle C in triangle ABC, specifically why it is assumed to be acute, and the determination of the value of sin C based on given values for angles A and B. The context includes trigonometric reasoning and potential misunderstandings regarding the nature of the triangle.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Some participants propose that angle C is acute because angle A is obtuse and angle B is acute, suggesting that a triangle can only have one obtuse angle.
  • Others argue that angle B is not a right angle, despite earlier claims, and question the classification of the triangle as a right triangle.
  • A participant expresses confusion regarding the quadrant system in relation to the triangle, noting that the triangle is not situated on a coordinate system.
  • One participant mentions the potential for the triangle to exist on a spherical surface, introducing an alternative perspective.
  • There is acknowledgment of previous discussions on this problem in another forum, indicating that it has been addressed before.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the classification of angle C or the nature of the triangle. There are competing views regarding the assumptions made about the angles and the triangle's properties.

Contextual Notes

There are unresolved issues regarding the definitions of angles and the classification of the triangle, as well as the implications of using trigonometric ratios in this context. The discussion reflects varying interpretations of the problem statement.

laprec
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Given that √5 tanA=-2 and CosB=8/17 in ∆ABC
State why we may assume that angle C is acute and determine the value of Sin CAttempt made:
tanA=-2/√5 CosB=8/17
A is obtuse angle of 138° or reflex angle 318.19°
B is an acute angle of61.9° or reflex angle298.1 °.Since it is a right triangle,therefore angle C is acute
Not sure the above is correct though!
Second Part: The diagram will be in the 4th quadrant since tan is negative and cos is positive
 

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laprec said:
Given that √5 tanA=-2 and CosB=8/17 in ∆ABC
State why we may assume that angle C is acute and determine the value of Sin CAttempt made:
tanA=-2/√5 CosB=8/17
A is obtuse angle of 138° or reflex angle 318.19°
Since tanA is negative, it is an obtuse angle. I get about 138.17 degrees. Since there can only be one obtuse angle in a triangle, angles B and C must be acute angles.

B is an acute angle of61.9° or reflex angle298.1 °.Since it is a right angle,therefore angle C is acute
What is a right angle? You have correctly calculated that B has measure 61.9 degrees so is NOT a right angle! At first I thought you might have meant that ABC is a right triangle but that is also not true (a right triangle cannot have an obtuse angle).

Not sure the above is correct though!
Second Part: The diagram will be in the 4th quadrant since tan is –ve and cos is +ve
I don't know what "diagram" you are talking about. The problem as you stated it has only a triangle, it is not on a coordinate system so there are NO quadrants.
 
Thanks a lot! I mean't to say it is a right triangle because I used Trig ratio. I am sorry the diagrams that were included in my attempt to solve the second part of the question are not displaying in the thread. I will make another attempt to attached the diagrams. Your efforts is highly appreciated.
 
I don’t think this triangle lies on a plane ... on a spherical surface, maybe?
 
Why is this posted here in the Challenges forum?
 
I moved it there due to the title. :o I've moved it back to Trigonometry.
 
This problem was asked, answered, and thanked by the OP on May 4 on another forum.
 

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