Consider an angle M with measure m≠90°, in a right triangle.
Let
OPP denote the length of the side opposite to M,
ADJ denote the length of the side adjacent to M, and
HYP denote the hypotenuse.
then:
Sin(M) = OPP/HYP
Cos(M)= ADJ/HYPP
Tan(M)=OPP/ADJ
Back to our problem,
using the Pythagorean we can find the length of AB:
[tex]|AB|^2+|AC|^2=|BC|^2\\\\|AB|^2+9^2=18^2\\\\|AB|^2=18^2-9^2\\\\|AB|^2=(18-9)(18+9)=9 \cdot 27=9 \cdot 9 \cdot3\\\\|AB|=9 \sqrt{3}
[/tex]
[tex]Sin(C)= \frac{OPP}{HYP}=\frac{9 \sqrt{3} }{18}= \frac{ \sqrt{3} }{2}[/tex]
[tex]Cos(B)= \frac{ADJ}{HYP}= \frac{9 \sqrt{3}}{18}= \frac{ \sqrt{3}}{2} [/tex]
[tex]Tan(C)= \frac{OPP}{ADJ}= \frac{9 \sqrt{3} }{9}= \sqrt{3} [/tex]
[tex]Sin(B)= \frac{OPP}{HYP}= \frac{9}{18}= \frac{1}{2} [/tex]
[tex]Tan(B)= \frac{OPP}{ADJ}= \frac{9}{9 \sqrt{3}}= \frac{1}{ \sqrt{3} }= \frac{ \sqrt{3}}{3} [/tex]
Answer: 1, 3, 4