Why Is the Sign Incorrect in My Solution for Tan A?

[odolwa99]

In this question, my answer to part (a) is correct, & leads into (b). With (b) my answer is also correct, except for the sign. Can anyone help me figure what I need to do?

Many thanks.

Homework Statement

Q. 90^{\circ}<A<180^{\circ} such that \sin(A+\frac{\pi}{6})+\sin(A-\frac{\pi}{6})=\frac{4\sqrt{3}}{5}. Find (a) \sin A & (b) \tan A

Homework Equations

The Attempt at a Solution

(a) \sin A\cos \frac{\pi}{6}+\cos A\sin{\pi}{6}+\sin A\cos\frac{\pi}{6}-\cos A\sin\frac{\pi}{6}=\frac{4\sqrt{3}}{5}
2\sin A\cos\frac{\pi}{6}=\frac{4\sqrt{3}}{5}
\frac{2\sqrt{3}\sin A}{2}=\frac{4\sqrt{3}}{5}
2\sqrt{3}\sin A=\frac{8\sqrt{3}}{10\sqrt{3}}
\sin A=\frac{4}{5}

(b) \tan A=\frac{\sin A}{\cos A}
\sin^2A+cos^2A=1
(\frac{4}{5})^2+cos^2A=1
\cos^2A=1-\frac{16}{25}
\cos A=\sqrt{\frac{9}{25}}
\cos A=\frac{3}{5}
\tan A=\frac{4}{5}/\frac{3}{5}
\tan A=\frac{4}{3}

Answer: (From textbook): (b) \frac{-4}{3}

 

[CAF123]

Use the condition given in the question: π/2 ≤ A ≤ π. if you like, draw the CAST diagram or sketch the cosine graph - what is the sign of cos here?
Note also \sqrt{9/25} = +/- 3/5

 

[odolwa99]

Great. Thank you.