# Trig ratio challenge

#### laprec

Kindly assist with this trig equation challenge:
Given that âˆš5 tanA=-2 and CosB=8/17 in âˆ†ABC
State why we may assume that angle C is acute and determine the value of Sin C

tanA=-2/âˆš5 CosB=8/17
A is obtuse angle of 138Â° or reflex angle 318.19Â°
B is an acute angle of61.9Â° or reflex angle298.1 Â°.Since it is a right angle,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 â€“ve and cos is +ve
Got stucked here!

#### mrtwhs

Kindly assist with this trig equation challenge:
Given that âˆš5 tanA=-2 and CosB=8/17 in âˆ†ABC
State why we may assume that angle C is acute and determine the value of Sin C

tanA=-2/âˆš5 CosB=8/17
A is obtuse angle of 138Â° or reflex angle 318.19Â°
B is an acute angle of61.9Â° or reflex angle298.1 Â°.Since it is a right angle,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 â€“ve and cos is +ve
Got stucked here!
What's all this 318.19 degrees and 298.1 degrees stuff? You said "in $$\displaystyle \triangle ABC$$". So all angles are less than 180 degrees. Angle A is obtuse so the other two angles are acute.

$$\displaystyle A+B+C=180^{\circ}$$ so $$\displaystyle C=180^{\circ}-(A+B)$$. Thus, $$\displaystyle \sin C=\sin(180^{\circ}-(A+B))=\sin(A+B)$$.

If $$\displaystyle \tan A = \dfrac{-2}{\sqrt{5}}$$ then $$\displaystyle \sin A=\dfrac{2}{3}$$ and $$\displaystyle \cos A = \dfrac{-\sqrt{5}}{3}$$

If $$\displaystyle \cos B = \dfrac{8}{17}$$ then $$\displaystyle \sin B=\dfrac{15}{17}$$.

See if you can finish it.

1 person

#### skipjack

Forum Staff
This seems to be a trick question.

Angle A is negative, and is about -41.81Â°.

sin(C) = Â±(16 + 15âˆš5)/51.

1 person

#### mrtwhs

Silly me for thinking that the problem was reasonable and didn't need to be checked.

$$\displaystyle \angle A \approx 138.19^{\circ}$$ and $$\displaystyle \angle B \approx 61.93^{\circ}$$. So there isn't much room for $$\displaystyle \angle C$$ in this triangle.

#### laprec

Thanks a million @skipjack and Mrtwhs! Your contributions has provided the needed insights to solve this problem. Once again Thank you!