May 4th, 2018, 09:32 AM  #1 
Newbie Joined: May 2018 From: South Africa Posts: 2 Thanks: 0  Trig ratio challenge
Kindly assist with this trig equation challenge: 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 C Attempt 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 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 –ve and cos is +ve Got stucked here! 
May 4th, 2018, 11:28 AM  #2  
Senior Member Joined: Feb 2010 Posts: 702 Thanks: 137  Quote:
$\displaystyle A+B+C=180^{\circ}$ 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.  
May 4th, 2018, 12:17 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,274 Thanks: 1959 
This seems to be a trick question. Angle A is negative, and is about 41.81°. Angle B is about ±61.93°. Angle C is about ±76.26°. sin(C) = ±(16 + 15√5)/51. 
May 4th, 2018, 12:46 PM  #4 
Senior Member Joined: Feb 2010 Posts: 702 Thanks: 137 
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. 
May 4th, 2018, 01:19 PM  #5 
Newbie Joined: May 2018 From: South Africa Posts: 2 Thanks: 0 
Thanks a million @skipjack and Mrtwhs! Your contributions has provided the needed insights to solve this problem. Once again Thank you!


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challenge, ratio, trig 
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