1. AB
AB is adjacent to the 50 deg angle.
You also know the hypotenuse, so use CAH.
[tex] \cos 50^\circ = \dfrac{AB}{9} [/tex]
[tex] AB = 9 \cos 50^\circ = 5.8 [/tex]
2. BC
BC is opposite the 50 deg angle.
You also know the hypotenuse, so use SOH.
[tex] \sin 50^\circ = \dfrac{BC}{9} [/tex]
[tex] BC = 9 \sin 50^\circ = 6.9 [/tex]
3. <C
<C is complementary to the 50 deg angle.
m<C + 50 = 90
m<C = 40
