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odolwa99
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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.
Q. [itex]90^{\circ}<A<180^{\circ}[/itex] such that [itex]\sin(A+\frac{\pi}{6})+\sin(A-\frac{\pi}{6})=\frac{4\sqrt{3}}{5}[/itex]. Find (a) [itex]\sin A[/itex] & (b) [itex]\tan A[/itex]
(a) [itex]\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}[/itex]
[itex]2\sin A\cos\frac{\pi}{6}=\frac{4\sqrt{3}}{5}[/itex]
[itex]\frac{2\sqrt{3}\sin A}{2}=\frac{4\sqrt{3}}{5}[/itex]
[itex]2\sqrt{3}\sin A=\frac{8\sqrt{3}}{10\sqrt{3}}[/itex]
[itex]\sin A=\frac{4}{5}[/itex]
(b) [itex]\tan A=\frac{\sin A}{\cos A}[/itex]
[itex]\sin^2A+cos^2A=1[/itex]
[itex](\frac{4}{5})^2+cos^2A=1[/itex]
[itex]\cos^2A=1-\frac{16}{25}[/itex]
[itex]\cos A=\sqrt{\frac{9}{25}}[/itex]
[itex]\cos A=\frac{3}{5}[/itex]
[itex]\tan A=\frac{4}{5}/\frac{3}{5}[/itex]
[itex]\tan A=\frac{4}{3}[/itex]
Answer: (From textbook): (b) [itex]\frac{-4}{3}[/itex]
Many thanks.
Homework Statement
Q. [itex]90^{\circ}<A<180^{\circ}[/itex] such that [itex]\sin(A+\frac{\pi}{6})+\sin(A-\frac{\pi}{6})=\frac{4\sqrt{3}}{5}[/itex]. Find (a) [itex]\sin A[/itex] & (b) [itex]\tan A[/itex]
Homework Equations
The Attempt at a Solution
(a) [itex]\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}[/itex]
[itex]2\sin A\cos\frac{\pi}{6}=\frac{4\sqrt{3}}{5}[/itex]
[itex]\frac{2\sqrt{3}\sin A}{2}=\frac{4\sqrt{3}}{5}[/itex]
[itex]2\sqrt{3}\sin A=\frac{8\sqrt{3}}{10\sqrt{3}}[/itex]
[itex]\sin A=\frac{4}{5}[/itex]
(b) [itex]\tan A=\frac{\sin A}{\cos A}[/itex]
[itex]\sin^2A+cos^2A=1[/itex]
[itex](\frac{4}{5})^2+cos^2A=1[/itex]
[itex]\cos^2A=1-\frac{16}{25}[/itex]
[itex]\cos A=\sqrt{\frac{9}{25}}[/itex]
[itex]\cos A=\frac{3}{5}[/itex]
[itex]\tan A=\frac{4}{5}/\frac{3}{5}[/itex]
[itex]\tan A=\frac{4}{3}[/itex]
Answer: (From textbook): (b) [itex]\frac{-4}{3}[/itex]