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

[laprec]

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

 

[HOI]

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.

 

[laprec]

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.

 

[skeeter]

I don’t think this triangle lies on a plane ... on a spherical surface, maybe?

 

[MarkFL]

Why is this posted here in the Challenges forum?

 

[Greg]

I moved it there due to the title. I've moved it back to Trigonometry.

 

[mrtwhs]

This problem was asked, answered, and thanked by the OP on May 4 on another forum.