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

AI Thread Summary
Angle C in triangle ABC is assumed to be acute because there can only be one obtuse angle in a triangle, and since angle A is obtuse at approximately 138 degrees, angles B and C must be acute. The value of angle B is calculated to be about 61.9 degrees, confirming it is acute. The discussion highlights confusion regarding the classification of the triangle and the relevance of quadrants, as the problem is strictly about triangle geometry, not coordinate systems. The original poster acknowledges previous discussions on this problem in another forum. The conversation emphasizes the importance of clarity in geometric definitions and the context of the problem.
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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