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

Attempt made:

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.