Question #50495

If cos(a)=-12/13 and cot(b)=24/7,90<a<180 and 180<b<270 then the quadrant in which a+b lies?
1

Expert's answer

2015-06-01T09:54:03-0400

Answer on Question #50495 – Math – Trigonometry

If cos(a)=12/13\cos(a) = -12/13 and cot(b)=24/7\cot(b) = 24/7, 90<a<18090{}^\circ < a < 180{}^\circ and 180<b<270180{}^\circ < b < 270{}^\circ then the quadrant in which a+ba + b lies?

Solution

Since cot(b)=cos(b)sin(b)\cot(b) = \frac{\cos(b)}{\sin(b)}, then 1+cot2(b)=1sin2(b)=1+24272=625491 + \cot^2(b) = \frac{1}{\sin^2(b)} = 1 + \frac{24^2}{7^2} = \frac{625}{49}, so sin2(b)=49625\sin^2(b) = \frac{49}{625}. Due to 180<b<270180{}^\circ < b < 270{}^\circ (so, sin(b)<0\sin(b) < 0 and cos(b)<0\cos(b) < 0), we obtain sin(b)=725\sin(b) = -\frac{7}{25} and cos(b)=1sin2(b)=149625=576625=2425\cos(b) = -\sqrt{1 - \sin^2(b)} = -\sqrt{1 - \frac{49}{625}} = -\sqrt{\frac{576}{625}} = -\frac{24}{25}.

Due to 90<a<18090{}^\circ < a < 180{}^\circ (so, sin(a)>0\sin(a) > 0) we obtain sin(a)=1cos2(a)=1144169=25169=513\sin(a) = \sqrt{1 - \cos^2(a)} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}.

Since sin(a+b)=sin(a)cos(b)+cos(a)sin(b)=5132425+1213725=120+842513=362513<0\sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b) = \frac{5}{13} \cdot \frac{-24}{25} + \frac{-12}{13} \cdot \frac{-7}{25} = \frac{-120 + 84}{25 \cdot 13} = \frac{-36}{25 \cdot 13} < 0 and cos(a+b)=cos(a)cos(b)sin(a)sin(b)=12132425513725=288+352513=3232513>0\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) = \frac{-12}{13} \cdot \frac{-24}{25} - \frac{5}{13} \cdot \frac{-7}{25} = \frac{288 + 35}{25 \cdot 13} = \frac{323}{25 \cdot 13} > 0 then we obtain that 270<a+b<360270{}^\circ < a + b < 360{}^\circ because there exists only one quadrant where x=a+bx = a + b, cosx>0\cos x > 0 and sinx<0\sin x < 0 simultaneously.

**Answer**: 270<a+b<360270{}^\circ < a + b < 360{}^\circ, the third quadrant.

www.AssignmentExpert.com


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS