第10題
平面上\(\overline{AC}=\overline{AD}\)，\(∠ABC=90^{\circ}\)，\(∠CAD=\alpha\)，\(∠CBD=\beta\)，\(∠CAB=\gamma\)，若\(\displaystyle cos\alpha=\frac{4}{5}\)，\(\displaystyle cos\beta=\frac{8}{17}\)，則\(tan\gamma=\)
。
[解答]
\(\begin{align}
& \cos \gamma =\frac{\overline{AB}}{\overline{AC}}=\frac{\overline{AB}}{\overline{AD}}=\frac{\sin \left( \frac{\pi }{2}-\beta -\left( \gamma -\alpha \right) \right)}{\sin \left( \frac{\pi }{2}+\beta \right)}=\frac{\cos \left( \alpha -\beta -\gamma \right)}{\cos \beta } \\
& \frac{8}{17}\cos \gamma =\cos \left( \alpha -\beta -\gamma \right) \\
& =\cos \left( \alpha -\beta \right)\cos \gamma +\sin \left( \alpha -\beta \right)\sin \gamma \\
& =\left( \cos \alpha \cos \beta +\sin \alpha \sin \beta \right)\cos \gamma +\left( \sin \alpha \cos \beta -\cos \alpha \sin \beta \right)\sin \gamma \\
& =\left( \frac{4}{5}\times \frac{8}{17}+\frac{3}{5}\times \frac{15}{17} \right)\cos \gamma +\left( \frac{3}{5}\times \frac{8}{17}-\frac{4}{5}\times \frac{15}{17} \right)\sin \gamma \\
& 37\cos \gamma =36\sin \gamma \\
& \tan \gamma =\frac{\sin \gamma }{\cos \gamma }=\frac{37}{36} \\
\end{align}\)
110.5.3補充
在\(xy\)平面上\(\overline{AC}=\overline{AD}\)，\(∠ABC=90^{\circ}\)，\(∠CAD=\alpha\)，\(∠CBD=\beta\)，\(∠CAB=\gamma\)，若\(\displaystyle sin\alpha=\frac{3}{5}\)，\(\displaystyle cos\beta=\frac{5}{13}\)，則\(tan\gamma=\)
。
(110彰化女中，
https://math.pro/db/thread-3514-1-1.html)
