Answer:
[tex]\huge\boxed{\sin\alpha=-\dfrac{16}{65},\ \tan\alpha=-\dfrac{16}{63}}\\\\or\\\\\huge\boxed{\sin\alpha=\dfrac{16}{65},\ \tan\alpha=\dfrac{16}{63}}[/tex]
Step-by-step explanation:
[tex]\cos\alpha=\dfrac{63}{65}\\\\\text{use}\ \sin^2\alpha+\cos^2\alpha=1\\\\\sin^2\alpha+\left(\dfrac{63}{65}\right)^2=1\\\\\sin^2\alpha+\dfrac{3969}{4225}=1\qquad\text{subtract}\ \dfrac{3969}{4225}\ \text{from both sides}\\\\\sin^2\alpha=\dfrac{4225}{4225}-\dfrac{3969}{4225}\\\\\sin^2\alpha=\dfrac{256}{4225}\to\sin\alpha=\pm\sqrt{\dfrac{256}{4225}}\\\\\sin\alpha=\pm\dfrac{\sqrt{256}}{\sqrt{4225}}\\\\\sin\alpha=\pm\dfrac{16}{65}[/tex]
[tex]\text{use}\ \tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}\\\\\text{substitute:}\\\\\tan\alpha=\dfrac{\pm\frac{16}{65}}{\frac{63}{65}}=\pm\dfrac{16}{65}\cdot\dfrac{65}{63}=\pm\dfrac{16}{63}[/tex]
