matematika wajib trigonometri mohon dibantu teman teman dan terima kasih

BC =
[tex] \sqrt{ {39}^{2} - {36}^{2} } [/tex]
BC =
[tex] \sqrt{225} = 15[/tex]
2a.
[tex] \sin( \beta ) = \frac{15}{39} = \frac{5}{13} [/tex]
b.
[tex] \cos( \beta ) = \frac{36}{39} = \frac{12}{13} [/tex]
c.
[tex] \tan( \beta ) = \frac{15}{36} = \frac{5}{12} [/tex]
d.
[tex] \csc( \beta ) = \frac{1}{ \sin( \beta ) } = \frac{13}{5} [/tex]
e.
[tex] \sec( \beta ) = \frac{1}{ \cos( \beta ) } = \frac{13}{12} [/tex]
f.
[tex] \cot( \beta ) = \frac{1}{ \tan( \beta ) } = \frac{12}{5} [/tex]
3.
1/2 + 1/2 = 1

4.
[tex] \tan(a - b) = \tan(60 - 30) \\ = \tan(30) = \frac{1}{3} \sqrt{3} [/tex]
5. sisi depan = 12
sisi samping = 6
sisi miring =
[tex] \sqrt{ {12}^{2} + {6}^{2} } = \sqrt{144 + 36} \\ = \sqrt{180} = 6 \sqrt{5} [/tex]
[tex] \sin( \alpha ) = \frac{12}{6 \sqrt{5} } = \frac{2 \sqrt{5} }{5} [/tex]
sin di kuadran 2 jadi nilainya positif
[tex] \cos( \alpha ) = - \frac{6}{6 \sqrt{5} } = - \frac{ \sqrt{5} }{5} [/tex]
cos bukan di kuadran 2 jadi nilainya negatif