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 May 4th, 2018, 08:32 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, 10:28 AM   #2
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 Originally Posted by laprec 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!
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.

 May 4th, 2018, 11:17 AM #3 Global Moderator   Joined: Dec 2006 Posts: 19,530 Thanks: 1750 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. Thanks from laprec
 May 4th, 2018, 11:46 AM #4 Senior Member     Joined: Feb 2010 Posts: 688 Thanks: 131 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, 12:19 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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